CHAPTER 9
Relationships Between Points,
Lines, and Planes
Review of Prerequisite Skills, p. 487
1. a. Yes; (2, 25) 5 (10, 212) 1 t(8, 27)
(2, 25) 5 (10, 212) 1 1(8, 27)
b. No; 12(1) 1 5(2) 2 13 5 9 2 0
c. Yes; (7, 23, 8) 5 (1, 0, 24) 1 t(2, 21, 4)
(7, 23, 8) 5 (1, 0, 24) 1 3(2, 21, 4)
d. No; (1, 0, 5) 5 (2, 1, 22) 1 t(4, 21, 2)
(21, 21, 7) 2 t(4, 21, 2)
There is no value of t that satisfies the equation.
2. Answers may vary. For example:
>
a. Vector: m 5 (7, 3) 2 (2, 5) 5 (5, 22)
>
r 5 (2, 5) 1 t(5, 22), tPR
Parametric: x 5 2 1 5t, y 5 5 2 2t, tPR
>
b. Vector: m 5 (4, 27) 2 (23, 7) 5 (7, 214)
>
r 5 (23, 7) 1 t(7, 214), tPR
Parametric: x 5 23 1 7t, y 5 7 2 14t, tPR
>
c. Vector: m 5 (23, 211) 2 (21, 0)
5 (22, 211)
>
r 5 (21, 0) 1 t(22, 211), tPR
Parametric: x 5 21 1 2 2t, y 5 211t, tPR
>
d. Vector: m 5 (6, 27, 0) 2 (1, 3, 5)
5 (5, 210, 25)
>
r 5 (1, 3, 5) 1 t(5, 210, 25), tPR
Parametric: x 5 1 1 5t, y 5 3 2 10t, z 5 5 2 5t,
tPR
>
e. Vector: m 5 (21, 5, 2) 2 (2, 0, 21)
5 (23, 5, 3)
>
r 5 (2, 0, 21) 1 t(23, 5, 3), tPR
Parametric: x 5 2 2 3t, y 5 25t, z 5 21 1 3t,
tPR
>
f. Vector: m 5 (12, 25, 27) 2 (2, 5, 21)
5 (10, 210, 26)
>
r 5 (2, 5, 21) 1 t(10, 210, 26), tPR
Parametric: x 5 2 1 10t, y 5 5 2 10t, z 5 21 2 6t,
tPR
>
3. a. Since n 5 (2, 6, 21), the Cartesian equation
of the plane is of the form 2x 1 6y 2 z 1 D 5 0,
where D is to be determined. Since P0 (4, 1, 23)
is on the plane, it must satisfy the equation. So
Calculus and Vectors Solutions Manual
2(4) 1 6(1) 2 1(23) 1 D 5 8 1 6 1 3 1 D
5 17 1 D 5 0. D 5 217, and the equation of the
plane is 2x 1 6y 2 z 2 17 5 0.
>
b. Since n 5 (0, 7, 0), the Cartesian equation of
the plane is of the form 7y 1 D 5 0, where D is to
be determined. Since P0 (22, 0, 5) is on the plane, it
must satisfy the equation. So 7(0) 1 D 5 0 1 D 5 0
thus D 5 0. The equation of the plane is
7y 5 0, or y 5 0.
>
c. Since n 5 (4, 23, 0), the Cartesian equation of
the plane is of the form 4x 2 3y 1 D 5 0, where
D is to be determined. Since P0 (3, 21, 22)
is on the plane, it must satisfy the equation. So
4(3) 2 3(21) 1 D 5 12 1 3 1 D 5 15 1 D 5 0.
D 5 215, and the equation of the plane is
4x 1 3y 2 15 5 0.
>
d. Since n 5 (6, 5, 23), the Cartesian equation of
the plane is of the form 6x 2 5y 1 3z 1 D 5 0,
where D is to be determined. Since P0 (0, 0, 0) is on
the plane, it must satisfy the equation. So
6(0) 2 5(0) 1 3(0) 1 D 5 0, or D 5 0. The
equation of the plane is 6x 2 5y 1 3z 5 0.
>
e. Since n 5 (11, 26, 0), the Cartesian equation
of the plane is of the form 11x 2 6y 1 D 5 0,
where D is to be determined. Since P0 (4, 1, 8)
is on the plane, it must satisfy the equation. So
11(4) 2 6(1) 1 D 5 44 2 6 1 D 5 38 1 D 5 0.
D 5 238, and the equation of the plane is
11x 2 6y 2 38 5 0.
>
f. Since n 5 (1, 1, 21), the Cartesian equation of
the plane is of the form x 1 y 1 2 z 1 D 5 0,
where D is to be determined. Since P0 (2, 5, 1)
is on the plane, it must satisfy the equation.
So 2 1 5 2 1 1 D 5 6 1 D 5 0. D 5 26, and the
equation of the plane is x 1 y 2 z 2 6 5 0.
4. Start by writing the given line in parametric
form: (x, y, z) 5 (2 1 s 1 2t, 1 2 s, 3s 2 5t), so
x 5 2 1 s 1 2t, y 5 1 2 s, and z 5 3s 2 5t.
Solving for s in each component, we get s 5 1 2 y
and substituting this into z 5 3s 2 5t gives
z 5 3(1 2 y) 2 5t 5 3 2 3y 2 5t.
9-1
>
3 2 3y 2 z
.
t5
5
So now 23 1 3y 1 z 5 25t and
Finally, substituting both equations for s and t into
x 5 2 1 s 1 2t, we get
x 5 2 1 (1 2 y) 1 2 a
3 2 3y 2 z
b.
5
>
Rearranging, we get
5x 5 10 1 5 2 5y 1 6 2 6y 2 2z
5x 1 11y 1 2z 2 21 5 0.
5. L1 is not parallel to the plane because (3, 0, 2) is
a point on the line and the plane. Substitute the
expressions for the components of the parametric
equation of the line into the equation of the plane.
4(3 1 t) 1 (22t) 2 (2 1 2t) 2 10 5 0
12 1 4t 2 2t 2 2 2 2t 2 10 5 0
050
This last statement is always true. So every point on
the line is also in the plane. Therefore, the line lies
on the plane.
For L2 substitute the expressions for the components
of the parametric equation of the line into the
equation of the plane.
4(23t) 1 (25 1 2t) 2 (210t) 2 10 5 0
212t 2 5 1 2t 1 10t 2 10 5 0
215 5 0
This last statement is never true. So the line and the
plane have no points in common. Therefore, L2 is
parallel to the plane. The line cannot lie on the
plane.
For L3 use the symmetric equation to rewrite x and
z in terms of y.
x 5 24y 2 23
z 5 2y 2 6
Substitute into the equation of the plane.
4(24y 2 23) 1 y 2 (2y 2 6) 2 10 5 0
216y 2 92 1 y 1 y 1 6 2 10 5 0
214y 2 96 5 0
This equation has a solution. Therefore, L3 and the
plane have a point in common and are not parallel.
However, (5, 27, 1) is a point that lies on the line
that does not lie on the plane. Therefore, L3 does
not lie in the plane.
6. a. A normal vector to this plane is determined by
calculating
the
cross product of the position vectors,
>
>
AB and AC
.
>
AB> 5 (2, 0, 0) 2 (1, 0, 21) 5 (1, 0, 1)
AC 5 (6, 21, 5) 2 (1, 0, 21) 5 (5, 21, 6)
9-2
>
AB 3 AC 5 ((0 ? 6) 2 (1 ? 21), (1 ? 5)
2 (1 ? 6), (1 ? 21) 2 (0 ? 5))
5 (0 1 1, 5 2 6, 21 2 0)
>
5 (1, 21, 21) 5 n .
If P(x, y, z) is any point on the plane, then
AP 5 (x 2 1, y, z 1 1), and if the normal to the
plane is (1, 21, 21), then
(x 2 1, y, z 1 1) ? (1, 21, 21) 5 0, so
x 2 1 2 y 2 z 2 1 5 0 and thus,
x 2 y> 2 z 2 2 5 0
b. PQ> 5 (6, 4, 0) 2 (4, 1, 22) 5 (2, 3, 2)
PR 5 (0, >0, 23)> 2 (4, 1, 22) 5 (24, 21, 21)
>
n 5 PQ 3 PR
5 (3(21) 2 2(21)), 2(24) 22(21),
2(21) 2 3(4))
5 (23 1 2, 28 1 2, 22 1 12) 5 (21, 26, 10)
Since (21, 26, 10) 5 21(1, 6, 210), we will use
(1, 6, 210) as the normal vector so that the coefficient
of x is positive.
If P(x, y, z) is any point on the plane,
>
then AP 5 (x 2 4, y 2 1, z 1 2), and if the normal
to the plane is (1, 6, 210), then
(x 2 4, y 2 1, z 1 2) ? (1, 6, 210) 5 0,
so x 2 4 1 6y 2 6 2 10z 2 20 5 0,
and thus x 1 6y 2 10z 2 30 5 0.
7. Answers may vary. For example: One direction
>
vector is m 5 (2, 21, 6) 2 (1, 24, 3) 5 (1, 3, 3).
Now we need to find a normal to the plane such
>
>
that n ? m 5 0. So (1, 3, 3) ? (a, 0, c) 5 0. Now we
have a 1 3c 5 0. A possible solution to this is
>
a 5 3, c 5 21. So n 5 (3, 0, 21) and the
Cartesian equation of the plane is 3x 2 z 5 0.
Since the plane is parallel to the y-axis, (0, 1, 0) is
another direction vector for the plane. Therefore, a
vector equation for the plane is
>
r 5 (1, 24, 3) 1 t(1, 3, 3) 1 s(0, 1, 0), s, tPR.
8. We are given the point A(21, 3, 4). We need to
>
find a normal vector n 5 (a, b, c) such that
a(x 1 1) 1 b(y 2 3) 1 c(z 2 4) 1 d 5 0.
The normal vector also must be perpendicular to the
two planes and their normals, (2, 21, 3) and
(5, 1, 23). One possible solution for the normal is
>
n 5 (0, 3, 1). So we have
3(y 2 3) 1 z 2 4 5 0
3y 1 z 2 9 2 4 5 0
And the equation of the plane is 3y 1 z 5 13.
Chapter 9: Relationships Between Points, Lines, and Planes
9.1 The Intersection of a Line
with a Plane and the Intersection
of Two Lines, pp. 496–498
1. a. First, show the parametric equations as
x 5 1 1 5s, y 5 2 1 s, z 5 23 1 s. Then the
plane can be written as p: x 2 2y 2 3z 5 6,
and the vector equation of the line is
>
r 5 (1, 2, 23) 1 s(5, 1, 1), sPR.
b. When we substitute the parametric equations into
the Cartesian equation for the plane, we get
(1 1 5s) 2 2(2 1 s) 2 3(23 1 s) 5 6
1 2 4 1 9 1 5s 2 2s 2 3s 5 6 2 0s 5 6
Note that by finishing the solution, we get 0s 5 0.
Since any real number will satisfy this equation, we
have an infinite number of solutions, and this line
lies on the plane.
2. a. A line and a plane can intersect in three ways:
Case 1: The line and the plane have zero points of
intersection. This occurs when the lines are not
incidental, meaning they do not intersect.
Case 2: The line and the plane have only one point
of intersection. This occurs when the line crosses
the plane at a single point.
Case 3: The line and the plane have an infinite
number of intersections. This occurs when the line
is coincident with the plane, meaning the line lies
on the plane.
b. Assume that the line and the plane have more
than one intersection, but not an infinite number.
For simplicity, assume two intersections. At the
first intersection, the line crosses the plane. In order
for the line to continue on, it must have the same
direction vector. If the line has already crossed the
plane, then it continues to move away from the
plane, and can not intersect again. So the line and
the plane can only intersect zero, one, or infinitely
many times.
>
3. a. The line r 5 s(1, 0, 0) is the x-axis.
>
b. The plane y 5 1 has the form r 5 (x, 1, z),
where x, and z are any values in R. So the plane is
parallel to the xz-plane, but just one unit away to
the right.
Calculus and Vectors Solutions Manual
c.
z
x
y
d. There are no intersections between the line and
the plane.
4. a. For x 1 4y 1 z 2 4 5 0, if we substitute the
parametric equations, we have
(22 1 t) 1 4(1 2 t)1 (2 1 3t) 1 4
5 22 1 4 1 2 1 t 2 4t 1 3t 2 4
5 0t 1 0
5 0. All values of t give a solution to the equation,
so all points on the line are also on the plane.
b. For the plane 2x 2 3y 1 4z 2 11 5 0, we can
substitute the parametric equations derived
>
from r 5 (1, 5, 6) 1 t(1, 22, 22):
x 5 1 1 t, y 5 5 2 2t, z 5 6 2 2t.
So we have 2(1 1 t)2 3(5 2 2t) 1 4(6 2 2t) 2 11
5 2 2 15 1 24 2 11 1 2t 1 6t 2 8t
5 0t 1 0
50
Similar to part a., all values of t give a solution to
this equation, so all points on the line are also on
the plane.
5. a. First, we should determine the parametric
equations from the vector form: x 5 21 2 s,
y 5 1 1 2s, z 5 2s. Substituting these into the
equation of the plane, we get
2(21 2 s) 2 2(1 1 2s) 1 3(2s) 2 1
5 22 2 2 2 1 2 2s 2 4s 1 6s
5 25 1 0s
Since there are no values of s such that 25 5 0, this
line and plane do not intersect.
b. Substituting the parametric equations into the
equation of the plane, we get
2(1 1 2t) 2 4(22 1 5t) 1 4(1 1 4t) 2 13
2 1 8 1 4 2 13 1 4t 2 20t 1 16t
5 1 1 0t
Since there are no values of t such that 1 5 0, there
are no solutions, and the plane and the line do not
intersect.
9-3
>
6. a. The direction vector is m 5 (21, 2, 2)
>
and the normal is n 5 (2, 22, 3), so if the line and
> >
the plane meet at right angles, m ? n 5 0. So
(21 ? 2) 1 (2 ? 22) 1 (2 ? 3) 5 22 2 4 1 6 5 0,
but 2(21) 2 2(1) 1 3(0) 2 1 5 25 2 0. So the
point on the line is not on the plane.
>
b. The direction vector is m 5 (2, 5, 4) and
>
the normal is n 5 (2, 24, 4), so if the line
> >
and the plane meet at right angles, m ? n 5 0. So
(2 ? 2) 1 (5 ? 24) 1 (4 ? 4) 5 4 2 20 1 16 5 0,
but 2(1) 2 4(22) 1 4(1) 2 13 5 1 2 0. So the
point on the line is not on the plane.
7. a. If the line and the plane intersect, then they are
equal at a particular point p. So we must substitute
the parametric equations into the equation of the
plane, and then solve for p.
(21 1 6p) 1 2(3 1 p) 2 (4 2 2p) 1 29
5 21 1 6 2 4 1 6p 1 2p 1 2p 1 29
5 30 1 10p
5 0. So now 210p 5 30 and p 5 23.
Now we must find the point at which the
line and plane intersect. To do this, just
substitute p 5 23 into the vector form of the line:
(21, 3, 4) 1 23(6, 1, 22) 5 (219, 0, 10).
b. If the line and the plane intersect, then they are
equal at a particular point p. So we must substitute
the parametric equations into the equation of the
plane, and then solve for p.
x 5 1 1 4s, y 5 22 2 s, z 5 3 1 s
2(1 1 4s) 1 7(22 2 s) 1 (3 1 s) 1 15
5 2 2 14 1 3 1 15 1 8s 2 7s 1 s
5 6 1 2s
5 0. So now 22s 5 6 and s 5 23. Now we must
find the point at which the line and plane intersect.
To do this, just substitute s 5 23 into the vector
form of the line:
(1, 22, 3) 1 23(4, 21, 1) 5 (211, 1, 0)
8. a. Comparing the x and y components in L1 and
L2, we have
3 1 4s 5 4 1 13t
1 2 s 5 1 2 5t
We can easily solve for one of the variables by
using the second equation: s 5 5t. Substituting this
back into the first equation: 3 1 20t 5 4 1 13t so
1 5 7t and thus t 5 17. So now we must solve for s:
5
3 1 4s 5 4 1 137 and s 5 20
28 5 7 . Placing these back
into the equations for L1 and L2:
9-4
5
41 2 45
L1: (3, 1, 5) 1 (4, 21, 2) 5 a , , b
7
7 7 7
13
5 5
41 2 5
L2: a4 1 , 1 2 , b 5 a , , b
7
7 7
7 7 7
The points must be equal for intersection to occur,
so there is no intersection and the lines are skew.
b. If we compare the z components of the two lines,
we see 2 5 8 2 6s or s 5 1. Substituting this back
into the x component (the y component would work
just as well), we have 3 1 m 5 23 1 7(1) 5 4, or
m 5 1. So now we can substitute m and s back into
the equations for the line, and we get
L3 : (3, 7, 2) 1 (1, 26, 0) 5 (4, 1, 2)
L4 : (23, 2, 8) 1 (7, 21, 26) 5 (4, 1, 2)
So (4, 1, 2) is the only point of intersection between
these two lines.
9. a. Comparing the y and z components of each
vector equation, we get the system of equations:
3 2 2p 5 3 2 2q
4 1 3p 5 24 1 11q
Note that from the first equation, p 5 q. So the
second equation becomes 4 1 3q 5 24 1 11q.
Solving for q, we get q 5 1. So from the earlier
relation, p 5 1. Placing these two values back into
the vector equations, we get
(22, 3, 4) 1 (6, 22, 3) 5 (4, 1, 7)
(22, 3, 24) 1 (6, 22, 11) 5 (4, 1, 7)
This shows that these two lines intersect at (4, 1, 7).
b. Comparing the x and z components of each
vector equation, we get the system of equations:
41r521s
6 1 4r 5 28 1 5s
Note that from the first equation, s 5 2 1 r. So the
second equation becomes 6 1 4r 5 2 1 5r.
Solving for r, we get r 5 4. So from the earlier
relation, s 5 6. Placing these two values back into
the vector equations, we get
(4, 1, 6) 1 4(1, 0, 4) 5 (8, 1, 22)
(2, 1, 28) 1 6(1, 0, 5) 5 (8, 1, 22)
This shows that these two lines intersect
at (8, 1, 22).
c. Comparing the x and z components of each
vector equation, we get the system of equations:
2 1 m 5 22 1 3p
11m512p
Note that from the second equation, m 5 2p. So
the first equation becomes 2 2 p 5 22 1 3p.
Solving for p, we get p 5 1. So from the earlier
relation, m 5 21. Placing these two values back
into the vector equations, we get
Chapter 9: Relationships Between Points, Lines, and Planes
(2, 2, 1) 2 (1, 1, 1) 5 (1, 1, 0)
(22, 2, 1) 1 (3, 21, 21) 5 (1, 1, 0)
This shows that these two lines intersect at (1, 1, 0).
d. Comparing the x and y components of each
vector equation, we get the system of equations:
1 1 0m 5 2 1 s
2 1 4m 5 3 2 2s
Note that from the first equation, s 5 21. So the
second equation becomes 2 1 4m 5 5.
Solving for m, we get m 5 34. Placing these two
values back into the vector equations, we get
3
21
(9, 1, 2) 2 (5, 0, 4) 5 a , 1, 21b
4
4
(8, 2, 3) 2 (4, 1, 22) 5 (4, 1, 5)
The two lines do not intersect, so they are skew.
10. At the point where the line intersects the z-axis,
the point Q(0, 0, q) equals the vector equation. So
for the x component, 23 1 3s 5 0 or s 5 1.
Substituting this into the vector equation, we get
(23, 2, 1) 1 (3, 22, 7) 5 (0, 0, 8). So q 5 8.
11. a. Comparing the x components, we get
22 1 7s 5 230 1 7t, which can be reduced to
28 1 7s 5 7t or s 2 t 5 4. Comparing the other
components, the same equation results.
b. From L1, we see that at (22, 3, 4), s 5 0. When
this occurs, t 5 4. Substituting this into L2, we get
(230, 11, 24) 1 4(7, 22, 2) 5 (22, 3, 4). Since
both of these lines have the same direction vector
and a common point, the lines are coincidental.
12. a. First, we must determine the values of s and t.
So comparing the x and z components, we get
23 1 s 5 1 2 3t
1 1 s 5 2 1 8t
From the second equation, s 5 1 1 8t. Substituting
this back into the first equation,
23 1 1 1 8t 5 1 2 3t or t 5 113 .
Substituting back into the second equation,
23 1 s 5 1 2 119 5 112 , and solving for s,
s 5 112 1 3 5 35
11 . Now we can solve for k. Compare
the y components after substituting s and t.
35
3
82
541 k
11
11
53 5 44 1 3k
or k 5 3.
b. The lines intersect when s 5 35
11 . The point of
35
35
intersection is ( 23 1 35
11 , 8 2 11 , 1 1 11 ) or
( 112 , 5311, 4611) .
Calculus and Vectors Solutions Manual
13. On the xz-plane, the point A has the coordinates
(x, 0, z), for any x, z. Similarly, on the yz-plane, the
point B has the coordinates (0, y, z) for any y, z. Now
the task is to find the required values of s for these
points. Starting with the x component of point B,
we have 0 5 28 1 2s or s 5 4. So point B is
(28, 26, 21) 1 4(2, 2, 1) 5 (0, 2, 3). For point A,
we need the y coordinate to equal 0. So 0 5 26 1 2s
or s 5 3. So point A is
(28, 26, 21) 1 3(2, 2, 1) 5 (22, 0, 2).
Now we need to find the distance.
d 5 "(0 2 (22))2 1 (2 2 0)2 1 (3 2 2)2
5 "4 1 4 1 1
5 "9
53
14. a. Comparing the y and z components of each
vector equation, we get the system of equations:
1 1 0p 5 21 2 2q
1 2 p 5 1 2 2q
Note that from the first equation, 2 5 22q or
q 5 21. So the second equation becomes
1 2 p 5 1 1 2 or p 5 22.
Placing these two values back into the vector
equations to find the intersection point A, we get
(2, 1, 1) 2 2(4, 0, 21) 5 (26, 1, 3)
(3, 21, 1) 2 (9, 22, 22) 5 (26, 1, 3)
Thus, the intersection point is (26, 1, 3).
b. A point on the xy plane has the form (x, y, 0). If
such a point is (26, 1, 0) then the distance from
this point is d 5 "0 1 0 1 32 5 3.
15. a. Comparing the x and y components of each
vector equation, we get the system of equations:
21 1 5s 5 4 1 0t
3 2 2s 5 21 1 2t
Note that from the first equation, 5 5 5s or s 5 1.
So the second equation becomes 3 2 2 5 21 1 2t
or t 5 1. Placing these two values back into the
vector equations to find the intersection point A,
we get
(21, 3, 2) 1 (5, 22, 10) 5 (4, 1, 12)
(4, 21, 1) 1 (0, 2, 11) 5 (4, 1, 12)
Thus, the intersection point is (4, 1, 12).
b. We need to find a vector (a, b, c) such that
5a 2 2b 1 10c 5 0
2b 1 11c 5 0
A possible solution to the second equation is
(a, 11, 22). If we substitute this into the first
equation, we get 5a 2 22 2 20 5 0 S 5a 5 42.
9-5
We can use this to get a solution of ( 425, 11, 22) . To
eliminate the fraction, we get (42, 55, 210). So the
>
vector equation is r 5 (4, 1, 12) 1 t(42, 55, 210),
tPR.
16. a.
z
L2
x
L1
y
b. The only point of intersection is at the origin
(0, 0, 0).
c. If p 5 0 and q 5 0, the intersection occurs at
(0, 0, 0).
17. a. Represent the lines parametrically, and
then substitute into the equation for the plane.
For the first equation, x 5 t, y 5 7 2 8t,
z 5 1 1 2t. Substituting into the plane equation,
2t 1 7 2 8t 1 3 1 6t 2 10 5 0. Simplifying,
0t 5 0. So the line lies on the plane.
For the second line, x 5 4 1 3s, y 5 21, z 5 1 2 2s
Substituting into the plane equation,
8 1 6s 2 1 1 3 2 6s 2 10 5 0. Simplifying,
0s 5 0. This line also lies on the plane.
b. Compare the x and y components:
4 1 3s 5 t
7 2 8t 5 21
From the second equation, t 5 1. Substituting back
into the first equation, 4 1 3s 5 1, or s 5 21.
Determine the point of intersection:
(1, 7 2 8, 1 1 2) 5 (1, 21, 3)
(4 2 3, 21, 1 1 2) 5 (1, 21, 3)
The point of intersection is (1, 21, 3).
18. Answers may vary. For example:
>
r 5 (2, 0, 0) 1 p(2, 0, 1), pPR
9.2 Systems of Equations, pp. 507–509
1. a. linear
b. not linear
c. linear
d. not linear
2. Answers may vary. For example:
x 1 y 1 2z 5 215
a. x 1 2y 1 z 5 23
2x 1 y 1 z 5 210
b. Subtract the first equation from the second, and
subtract twice the first equation from the third.
9-6
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 2 y 2 3z 5 20
Add the second and third equation.
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 1 0y 2 4z 5 32
From the third equation, z 5 28.
Substitute z back into the second equation,
2y 2 8 5 212
2y 5 212 1 8 5 24
So y 5 4. Now substitute y and z back into the first
equation.
x 1 4 1 2(28) 5 x 2 12 5 215
And so x 5 23. Thus the solution is (23, 4, 28)
as expected.
3
3. a. 27 2 3(5) 1 4 a b 5 27 2 15 1 3 5 219
4
3
27 2 8 a b 5 27 2 6 5 213
4
27 1 2(5) 5 3
Yes, ( 27, 5, 34) is a solution.
b.
3
3(27) 2 2(5) 1 16 a b 5 221 2 10 1 12 5 219
4
3(27) 2 2(5) 5 221 2 10 5 231
2 223
3
8(27) 2 5 1 4 a b 5 256 2 5 1 3 5 258
4
Because the second equation fails to produce an
equality, ( 27, 5, 34) is not a solution.
4. a. Solve for y. y 5 23
The solution is (22, 23).
b. Multiply the second equation by 6
3x 1 5y 5 221
x 2 3y 5 7
Add 3 times the first equation to 5 times the second
equation.
3x 1 5y 5 221
14x 5 228
From the second equation, x 5 22.
Substituting x back into the first equation,
3(22) 1 5y 5 221
5y 5 215
So y 5 23.
The two systems are equivalent because they have
the same solution.
5. a. Add the second equation to 5 times the first
equation.
Chapter 9: Relationships Between Points, Lines, and Planes
2x 2 y 5 11
11x 5 66
Solve for x in the second equation, x 5 6. Substitute x
back into the first equation
2(6) 2 y 5 11
2y 5 11 2 12 5 21
So y 5 1
Therefore, the solution is (6, 1).
b. Subtract three times the first equation from twice
the second equation.
2x 1 5y 5 19
0x 2 7y 5 235
From the second equation, y 5 5.
Substitute y back into the first equation.
2x 1 5(5) 5 19
2x 5 19 2 25 5 26
So x 5 23
Therefore, the solution is (23, 5).
c. Add the second equation to 3 times the first
equation to the second equation
2x 1 2y 5 10
0x 1 11y 5 33
From the second equation, y 5 3.
Substitute y back into the first equation.
2x 1 2(3) 5 10
2x 5 4
So x 5 24.
Therefore the solution is (24, 3).
6. a. These two lines are parallel, and therefore
cannot have an intersection.
b. The second equation is five times the first,
therefore the lines are coincident.
7. a. Let x 5 t. So 2t 2 y 5 3 then y 5 2t 2 3.
b. Let x 5 t, y 5 s. Then t 2 2s 1 z 5 0 and
z 5 2s 1 t.
8. a. If x 5 t, y 5 22t 2 11, then y 5 22x 2 11
and so 2x 1 y 5 211 is the required linear equation.
b. 2x 1 y 5 211
2(3t 1 3) 1 (26t 2 17) 5 6t 2 6t 1 6 2 17
5 211
9. a. The two equations will have no solutions when
k 2 12, since they will be parallel should this occur.
b. It is impossible to have only one solution for these
two equations. They have exactly the same direction
vector. They will never intersect at exactly one place.
c. The two equations will have infinitely many
solutions when k 5 12. When this occurs, the two
equations are coincident.
Calculus and Vectors Solutions Manual
10. a. There are infinitely many solutions to this
equation. This is reason why it is represented
graphically as a line.
b. Let x 5 t. So 2t 1 4y 5 11, then 4y 5 11 2 2t
and y 5 114 2 12 t, tPR
c. This equation will not have any integer solutions
because the left hand side is an even function and
the right side is an odd function.
11. a. Add the second equation to 22 times the
first.
x 1 3y 5 a
0x 2 3y 5 b 2 2a
Divide the second equation by 23 to get
y 5 2 13 b 1 23 a. Now substitute this back into
the first equation.
1
2
x 1 3a2 b 1 ab 5 a
3
3
x 2 b 1 2a 5 a
x 5 2a 1 b
b. Since they have different direction vectors, these
two equations are not parallel or coincident and will
intersect somewhere.
12. a. Add the third equation to the first to eliminate z.
x1y1z50
x 2 y 1 0z 5 1
x 1 2y 1 0z 5 25
Add twice the second equation to the third equation
to eliminate
Add twice the second equation to the third equation
to eliminate y.
x1y1z50
x 2 y 1 0z 5 1
3x 1 0y 1 0z 5 23
Divide the third equation by 23 to get x 5 21.
Now substitute into the second equation.
21 2 y 5 1
y 5 22
Finally, substitute x and y to get
21 1 22 1 z 5 0
So z 5 3. Therefore, the solution is (21, 22, 3).
b. Add the first equation to 22 times the second,
and add the first equation to 22 times the third.
2x 2 3y 1 z 5 6
0x 2 5y 2 3z 5 256
0x 2 y 1 3z 5 40
Now add the second equation to 21 times the third.
2x 2 3y 1 z 5 6
0x 2 5y 2 3z 5 256
0x 2 4y 1 0z 5 216
9-7
From the third equation, y 5 4.
Now substitute this into the second equation.
25(4) 2 3z 5 256
23z 5 236
z 5 12
Now substitute these two values back into the first
equation.
2x 2 3(4) 1 12 5 6
2x 5 6, x 5 3
So the solution is (3, 4, 12).
c. Add the second equation to 21 times the third.
x 1 y 1 0z 5 10
0x 1 y 1 z 5 22
2x 1 y 1 0z 5 2
Add the third equation to the first equation.
x 1 y 1 0z 5 10
0x 1 y 1 z 5 22
0x 1 2y 1 0z 5 12
So y 5 6. Now substitute into the other two
equations.
x 1 6 5 10 S x 5 4
6 1 z 5 22 S z 5 28
So the solution is (4, 6, 28).
d. To eliminate fractions, multiply each of the
equations by 60.
20x 1 15y 1 12z 5 840
15x 1 12y 1 20z 5 21260
12x 1 20y 1 15z 5 420
Add 3 times the first equation to 24 times the
second, and add 3 times the first equation to 25
times the third.
20x 1 15y 1 12z 5 840
0x 2 3y 2 44z 5 7560
0x 2 55y 2 39z 5 420
Now add 55 times the second equation to 23 times
the third equation.
20x 1 15y 1 12z 5 840
0x 2 3y 2 44z 5 7560
0x 1 0y 2 2303z 5 414 540
Divide the third equation through by 22303 to get
z 5 2180. Substituting z back into the second
equation.
23y 2 44(2180) 5 7560 S 23y 5 2360
So y 5 120. Now substitute these two values back
into the first equation.
20x 1 15(120) 1 12(2180) 5 840
20x 5 840 2 1800 1 2160 5 1200
So x 5 60. Therefore the solution is (60, 120, 2180).
9-8
e. Note that if 2x 2 y 5 0 S y 5 2x, and
2z 2 x 5 0 S z 5 12 x. So we substitute these two
relations into the second equation.
1
7
2(2x) 2 x 5 x 5 7 S x 5 2
2
2
So now z 5 1, y 5 4, and the solution is (2, 4, 1).
f. Add the first equation to 22 times the second
equation.
x 1 y 1 2z 5 13
22x 1 0y 2 7z 5 238
2x 1 0y 1 6z 5 32
Add the second and third equations.
x 1 y 1 2z 5 13
22x 1 0y 2 7z 5 238
0x 1 0y 2 z 5 26
So from the third equation, z 5 6.
Substituting into the second equation,
22x 2 42 5 238
22x 5 4 S x 5 22
Finally, substituting both values into the first
equation,
22 1 y 1 12 5 13 S y 5 3.
So the final solution is (22, 3, 6).
13. Answers may vary. For example:
a.
Three lines parallel
z
L3
L2
L1
y
x
Two lines coincident
and the third parallel
z
L3
L2
L1
y
x
Chapter 9: Relationships Between Points, Lines, and Planes
Two parallel lines cut
by the third line
z
L3
L2
L1
y
x
The lines form a triangle
z
L2
L3
b.
y
L1
x
Lines meet in a point
z
L1
L3
L2
y
x
c.
Three coincident lines
z
L1
L2
L3
y
x
14. a. Add 21 times the first equation and the
second equation. Add 21 times the first equation
and the third equation.
x1y1z5a
0x 1 0y 2 z 5 b 2 a
2x 1 0y 1 0z 5 c 2 a
Calculus and Vectors Solutions Manual
So z 5 a 2 b, x 5 a 2 c. Then substitute into the
first equation.
a2c1y1a2b5a
y 5 2a 1 b 1 c
So the final solution is (a 2 c, 2a 1 b 1 c, a 2 b).
15. a. For two equations to have no solutions, they
must be parallel—meaning it must have a the same
direction vector. So if k 5 2, then the lines are
parallel.
b. If two equations have an infinite number of
solutions, then the lines must be coincident. One
way to do this is if the second equation is a multiple
of the first equation. To achieve this, k 5 22.
c. For two equations to have a unique solution, they
must have only one intersection. From a., we saw
that if k 5 2, the lines are parallel, and if k 5 22,
then they are coincident. Since the only other option
is for the lines to have a unique solution, k 2 6 2.
9.3 The Intersection of Two Planes,
pp. 516–517
1. a. This means that the two equations represent
planes that are parallel and not coincident.
b. Answers may vary. For example: x 2 y 1 z 5 1,
x 2 y 1 z 5 22
2. a. The solution to the system of equations is:
x 5 12 1 12 s 2 t, y 5 s, z 5 t, s, tPR. The two
planes are coincident.
b. Answers may vary. For example:
x 2 y 1 z 1 21; 2x 2 2y 1 2z 5 22
3. a. 2z 5 24 1 z 5 22.
x 2 y 1 (22) 1 21
x 2 y 1 1.
x 5 1 1 s, y 5 s, z 5 22, sPR
The two planes intersect in a line.
b. Answers may vary. For example:
x 2 y 1 z 5 21; x 2 y 2 z 5 3.
4. a. 1 2x 1 y 1 6z 5 p; 2 x 1 my 1 3z 5 q
For the planes to be coincident equation 2 must be
a multiple of equation 1 . Since the coefficients of
x and z in equation 1 are twice that of the x and z
coefficients in equation 2 all of the coefficients
and constants in equation 2 must be half of the
corresponding coefficients in equation 1 . So:
1
m 5 , p 5 2q. q 5 1, and p 5 2.
2
The value for m is unique, but p just has to be twice
q and arbitrary values can be chosen.
9-9
b. For parallel planes all of the coefficients of the
variables must be multiples of each other, but the
constant terms must differ by a different constant.
So a possible solution is:
1
m 5 , q 5 1, and p 5 3.
2
The value for m is again unique but p and q can be
arbitrarily chosen as long as p 2 2q.
c. For the two planes to intersect at right angles the
>
two normal vectors, n1 5 (2, 1, 6) and
>
n2 5 (1, m, 3), must satisfy:
>
>
n1 ? n2 5 0.
>
>
n1 ? n2 5 2 1 m 1 18 5 0
m 5 220. This value is unique, since only one
value was found to satisfy the given conditions.
d. From c. we know that in order to intersect in
right angles m 5 220. Choose p 5 1, q 5 1.
The value for m is unique from the solution to c.,
but the values for p and q can be arbitrary since the
only value which can change the angle between the
planes is m.
5. a. Letting z 5 s:
y 5 23s.
x 1 2(23s) 2 3s 5 0.
x 5 9s
The solution is:
x 5 9s, y 5 23s, z 5 s, sPR
b. Letting y 5 t.
t 1 3z 5 0
3z 5 2t
1
z 5 2 t.
3
1
x 1 2t 2 3a2 bt 5 0
3
x 1 3t 5 0
x 5 23t.
The solution is:
1
x 5 23t, y 5 t, z 5 2 t; tPR.
3
c. Since t is an arbitrary real number we can
express t as:
t 5 23s; sPR.
Substituting this into the solution for b. shows that
the two solutions are equivalent.
6. a. Equation 2 is twice that of equation 1 , so
they represent intersecting coincident planes.
b. The coefficients of each variable are the same,
but the constant terms are different, so the equations
represent non-intersecting parallel planes.
9-10
c. The coefficients of the x and z variables are the
same but the y coefficients are different. So the
equations represent planes that intersect in a line.
d. The coefficients of each variable from equation 1
to 2 are not the same multiple. Therefore the
equations represent planes that intersect in a line.
e. The intersection is a line by the same reasoning
as d.
f. The intersection is a line by the same reasoning as d.
7. a. x 5 1 2 s 2 t, y 5 s, z 5 t, s, tPR
b. There is no solution since the planes are parallel.
c. 1 2 2 :
22y 5 4
y 5 22.
x 2 2 1 2z 5 22
x 1 2z 5 0
x 5 22z.
x 5 22s, y 5 22, z 5 s, sPR.
d. Let z 5 s; sPR.
From 2 :
x 5 y 1 6.
(y 1 6) 1 y 1 2s 5 4
2y 1 2s 5 22
y 5 2s 2 1.
x 5 2s 1 5, y 5 2s 2 1, z 5 s, sPR.
e. 22 ? 2 : 2x 2 4y 2 2z 5 22
Adding 1 :
4x 2 5y 5 0.
5
x 5 y.
4
Let y 5 s, sPR.
5
2a sb 2 s 1 2z 5 2
4
3
s 1 2z 5 2
2
3
z 5 1 2 s.
4
3
5
x 5 s, y 5 s, z 5 1 2 s, sPR
4
4
f. x 2 y 1 2(4) 5 0
x 5 y 2 8.
x 5 s 2 8, y 5 s, z 5 4, sPR.
8. a. The system will have an infinite number of
solutions for any value of k. When k 5 2 equation
2 will be twice that of 1 so the solution is a plane:
x 5 1 2 s 2 2t, y 5 s, z 5 t, s, tPR.
For any other value of k the solution will be a line.
For example k 5 0:
2y 5 24z
y 5 22z.
Chapter 9: Relationships Between Points, Lines, and Planes
x 1 (22z) 1 2z 5 1
x 5 1.
x 5 1, y 5 22s, z 5 s, sPR.
b. No there is no value of k for which the system will
not have a solution. The only time when there is no
solution is when the corresponding coefficients for
each variable differ by a common multiple between
equations, and the constant terms differ by a different
multiple. The only way the first condition is satisfied is
when k 5 2, but when this happens the constant terms
differ by the same factor as the variables, namely 2.
9. The line of intersection of the two planes:
p1: 2x 2 y 1 z 5 0, p2: y 1 4z 5 0 is:
y 5 24z
2x 2 (24z) 1 z 5 0
2x 5 25z
5
x 5 2 z.
2
5
x 5 2 s, y 5 24s, z 5 s, sPR.
2
The direction vector is ( 2 52, 24, 1) or (25, 28, 2).
>
r1 5 s(25, 28, 2), sPR. Since the line we are
looking for is parallel to this line, we know that the
direction vector must be the same. The line passes
through (22, 3, 6) and has direction vector
(25, 28, 2). The equation of the line is
>
r2 5 (22, 3, 6) 1 s(25, 28, 2), sPR.
10. The line of intersection of the two planes,
2x 2 y 1 2z 5 0 and 2x 1 y 1 6z 5 4 is:
4x 1 8z 5 4
x 5 1 2 2z.
2(1 2 2z) 2 y 1 2z 5 0
2 2 y 2 2z 5 0
y 5 2 2 2z.
x 5 1 2 2s, y 5 2 2 2s, z 5 s, sPR.
In order for the a line to be contained in the plane
we need to check that the values for x, y, and z
always satisfy the plane equation:
5x 1 3y 1 16z 2 11 5 0.
5(1 2 2s) 1 3(2 2 2s) 1 16(s) 2 11 5 0
5 1 6 2 11 2 10s 2 6s 1 16s 5 0
0 5 0. Since this is true the line is contained in the
plane.
11. a. p1: 2x 1 y 2 3z 5 3, p2: x 2 2y 1 z 5 21.
p1 2 2p2: 5y 2 5z 5 5
y 5 1 1 z.
2x 1 (1 1 z) 2 3z 5 3
2x 2 2z 5 2
x 5 1 1 z.
x 5 1 1 s, y 5 1 1 s, z 5 s, sPR.
Calculus and Vectors Solutions Manual
b. L meets the xy-plane when z 5 0.
x 5 1, y 5 1. A 5 (1, 1, 0).
L meets the z-axis when both x and y are zero:
s 5 21.
z 5 21.
B 5 (0, 0, 21)
The length of AB is therefore:
"12 1 12 1 12 5 "3 or about 1.73.
12. The line with equation x 5 22y 5 3z has
parametric equations: x 5 s, y 5 2 12 s, z 5 13 s, sPR.
This has the equivalent vector form:
1 1
>
r 5 sa1, 2 , b, sPR.
2 3
The line of intersection of the two planes
x 2 y 1 z 5 1 and 2y 2 z 5 0 is:
1
y5 z
2
1
x2 z1z51
2
1
x 5 1 2 z.
2
1
1
x 5 1 2 2 t, y 5 2 t, z 5 t, tPR. Which has a vector
equation
of:
>
r 5 (1, 0, 0) 1 t (2 12, 12, 1), tPR. The vector
equation of the plane with the given properties is
thus:
1 1
1 1
>
r 5 (1, 0, 0) 1 t a2 , , 1b 1 s a1, 2 , b, s, tPR.
2 2
2 3
The normal vector for the plane is then:
1 1
1 1
1
1 1
a2 , , 1b 3 a1, 2 , b 5 a ? b 2 a1 ? 2 b,
2 2
2 3
2 3
2
1 1
1 1
1
2 7 1
1 ? 1 2 a2 ? b, 2 a2 b 2 ? 1 5 a , , 2 b.
2 3
2 2
2
3 6 4
Or equivalently (8, 14, 23).
The Cartesian equation is then:
8x 1 14y 2 3z 1 D 5 0, and must contain the
point (1, 0, 0).
8(1) 1 D 5 0.
D 5 28.
8x 1 14y 2 3z 2 8 5 0.
Mid-Chapter Review, pp. 518–519
>
1. a. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 15 1 5t
t 5 23
9-11
x 5 4 1 2(23), y 5 23 2 3(23),
z 5 15 1 5(23)
x 5 22, y 5 6, z 5 0
(22, 6, 0)
>
b. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 23 2 3t
t 5 21
x 5 4 1 2(21), y 5 23 2 3(21),
z 5 15 1 5(21)
x 5 2, y 5 0, z 5 10
(2, 0, 10)
>
c. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 4 1 2t
t 5 22
x 5 4 1 2(22), y 5 23 2 3(22),
z 5 15 1 5(22)
x 5 0, y 5 3, z 5 5
(0, 3, 5)
2. a.–e. Answers may vary. For example:
A(2, 1, 3), B(3, 22, 5), C(28, 25, 7)
a 5 (22.5, 23.5, 6)
b 5 (23, 22, 5)
c 5 (2.5, 20.5, 4)
m1 5 (Aa) 5 (24.5, 24.5, 3) 5 (3, 3, 22)
m2 5 (Bb) 5 (26, 0, 0) 5 (1, 0, 0)
m3 5 (Cc) 5 (10.5, 4.5, 23) 5 (7, 3, 22)
Then substitute in the point and the direction vector
to find the equation of the line.
A(2, 1, 3), B(3, 22, 5), C(28, 25, 7)
m1 5 (Aa) 5 (24.5, 24.5, 3) 5 (3, 3, 22)
m2 5 (Bb) 5 (26, 0, 0) 5 (1, 0, 0)
m3 5 (Cc) 5 (10.5, 4.5, 23) 5 (7, 3, 22)
>
A: r 5 (2, 1, 3) 1 t(3, 3, 22), tPR
x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
>
B: r 5 (3, 22, 5) 1 t(1, 0, 0), tPR
x 5 3 1 t, y 5 22, z 5 5, tPR
>
C: r 5 (28, 25, 7) 1 t(7, 3, 22), tPR
x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
A: x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
B: x 5 3 1t, y 5 22, z 5 5, tPR
C: x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
y 5 22 5 1 1 3t
t 5 21
x 5 2 1 3(21), y 5 1 1 3(21),
z 5 3 2 2(21)
x 5 21, y 5 22, z 5 5
(21, 22, 5)
9-12
A: x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
B: x 5 3 1 t, y 5 22, z 5 5, tPR
C: x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
y 5 22 5 25 1 3t
t51
x 5 28 1 7(1), y 5 25 1 3(1), z 5 7 2 2(1)
x 5 21, y 5 22, z 5 5
(21, 22, 5)
The three medians meet at (21, 22, 5).
3. a. L1: 5x 1 y 1 2z 1 15 5 0
L2: 4x 1 y 1 2z 1 8 5 0
L1 2 L2: x 1 7 5 0
So x 5 27.
L1: y 1 2z 2 20 5 0
L2: y 1 2z 2 20 5 0
z 5 t,
y 1 2(t) 2 20 5 0
y 5 20 2 2t
>
r 5 (27, 20, 0) 1 t(0, 22, 1), tPR
b.
L1: 4x 1 3y 1 3z 2 2 5 0
L2: 5x 1 2y 1 3z 1 5 5 0
2L1 2 3L2: 27x 2 3z 2 19 5 0
z 5 7t,
27x 2 3(7t) 2 19 5 0,
19
x 5 23t 2
7
19
4a23t 2 b 1 3y 1 3(7t) 2 2 5 0
7
30
y 5 23t 1
7
19 30
>
r 5 a2 , , 0b 1 t(3, 3, 27), tPR
7 7
>
c. L1: r 5 (27, 20, 0) 1 t(0, 22, 1), tPR
19 30
>
L2: r 5 a2 , , 0b 1 t(3, 3, 27), tPR
7 7
L1: x 5 27, y 5 20 2 2t, z 5 t
19
30
L2: x 5 2 1 3t, y 5
1 3t, z 5 27t
7
7
30
19
2 1 3t 5 27, t 5 2
7
21
19
30
30
30
x 5 2 1 3a2 b, y 5
1 3a2 b,
7
21
7
21
30
z 5 27a2 b
21
x 5 27, y 5 0, z 5 10
(27, 0, 10)
4. a. p1: 3x 1 y 1 7z 1 3 5 0
p2: x 2 13y 2 3z 2 38 5 0
Chapter 9: Relationships Between Points, Lines, and Planes
13p1 1 p2: 40x 1 88z 1 1 5 0
z 5 t,
40x 1 88(t) 1 1 5 0
11t
1
x52
2
5
40
11t
1
3a2
2 b 1 y 1 7(t) 1 3 5 0
5
40
2t
117
y52 2
5
40
2t
11t
1
117
x52
2 ,y52 2
, z 5 t, tPR
5
40
5
40
p1: x 2 3y 1 z 1 11 5 0
b.
p2: 6x 2 13y 1 8z 2 28 5 0
26p1 1 p2: 5y 1 2z 2 94 5 0
z 5 s,
5y 1 2(s) 2 94 5 0
2
94
y52 s1
5
5
2
94
x 2 3a2 s 1 b 1 (s) 1 11 5 0
5
5
227
11
x52 s1
5
5
2
11
227
94
x52 s1
, y 5 2 s 1 , z 5 s, sPR
5
5
5
5
c. The lines found in 4. a. and 4. b. do not intersect,
because they are in parallel planes.
5. a. For there to be no solution the lines must be
inconsistent with each other.
L1: x 1 ay 5 9
L2: ax 1 9y 5 227
1
a
5
a
9
a 5 63
For a 5 3:
L1: x 1 3y 5 9
L2: 3x 1 9y 5 227
For a 5 23, the equations are equivalent.
So there is no solution when a 5 3.
b. To have an infinite number of solutions, the lines
must be proportional.
L1: x 1 ay 5 9
L2: ax 1 9y 5 227
23(x 1 ay 5 9) 5 23x 2 3ay 5 227
L1: 23x 2 3ay 5 227
L2: ax 1 9y 5 227
a 5 23
c. The system has one solution when a 2 3 or
a 2 23, because other values lead to an infinite
number of solutions or no solution.
Calculus and Vectors Solutions Manual
x 2 11
y24
z 2 27
5
5
5s
2
24
5
L2: x 5 0, y 5 1 2 3t, z 5 3 1 2t, tPR
L1: x 5 2s 1 11, y 5 24s 1 4, z 5 27 1 5s
x 5 0 5 2s 1 11,
s 5 25.5
y 5 24(25.5) 1 4, z 5 27 1 5(25.5)
x 5 0, y 5 26, z 5 20.5
25
y 5 26 5 1 2 3t, t 5 2
3
7
z 5 20.5 5 3 1 2t, t 5 2
4
Since there is no t-value that satisfies the equations,
there is no intersection, and these lines are skew.
x25
z14
7. a. L1:
5y225
5s
2
23
L2: (x 2 3, y 2 20, z 2 7) 5 t(2, 24, 5), tPR
L1: x 5 2s 1 5, y 5 s 1 2, z 5 23s 2 4
L2: x 5 2t 1 3, y 5 24t 1 20, z 5 5t 1 7
x 5 2t 1 3 5 2s 1 5
y 5 s 1 2 5 24t 1 20
z 5 23s 2 4 5 5t 1 7
L3: 2t 2 2s 2 2 5 0
L4: 4t 1 s 2 18 5 0
L5: 5t 1 3s 1 11 5 0
L3 1 2L4: 10t 2 38 5 0, t 5 3.8
3L3 1 2L5: 16t 1 16 5 0, t 5 21
b. Since there is no t-value that satisfies the
equations, there is no intersection, and these lines
are skew.
8. L1: x 5 1 1 2s, y 5 4 2 s, z 5 23s, sPR
L2: x 5 23, y 5 t 1 3, z 5 2t, tPR
x 5 23 5 1 1 2s
s 5 22
x 5 23, y 5 6, z 5 6
(23, 6, 6)
>
9. a. L1: r 5 (5, 1, 7) 1 s(2, 0, 5), sPR
>
L2: r 5 (21, 21, 3) 1 t(4, 2, 21), tPR
L1: x 5 5 1 2s, y 5 1, z 5 7 1 5s
L2: x 5 21 1 4t, y 5 21 1 2t, z 5 3 2 t
y 5 1 5 21 1 2t,
t51
x 5 21 1 4(1), y 5 21 1 2(1),
z 5 3 2 (1)
x 5 3, y 5 1, z 5 2
(3, 1, 2)
>
b. L1: r 5 (2, 21, 3) 1 s(5, 21, 6), sPR
>
L2: r 5 (28, 1, 29) 1 t(5, 21, 6), tPR
These lines are the same, so either one of these
lines can be used as their intersection.
6. L1:
9-13
10. a. Answers may vary. For example:
i. coincident
z
L1
L2
y
x
ii. parallel and distinct
z
L2
L1
y
x
iii. skew
z
L1
x
y
L2
iv. intersect in a point
z
L1
L2
y
x
b. i. When lines are the same, they are a multiple
of each other.
ii. When lines are parallel, one equation is a
multiple of the other equation, except for the
constant term.
9-14
iii. When lines are skew, there are no common
solutions to make each equation consistent.
iv. When the solution meets in a point, there is only
one unique solution for the system.
11. a. A line and plane have an infinite number of
points of intersection when the line lies in the plane.
b. Answers may vary. For example:
>
r 5 t(3, 25, 23), tPR
>
r 5 t(3, 25, 23) 1 s(1, 1, 1), t, sPR
12. a. 1 2x 1 3y 5 30
2
x 2 2y 5 213
Equation 1 2 (2 3 equation 2 ): 7y 5 56
y58
2x 1 24 5 30
x53
(3, 8)
b. 1
x 1 4y 2 3z 1 6 5 0
2 2x 1 8y 2 6z 1 11 5 0
There is no solution to this system, because the
planes are parallel, but one plane lies above the
other.
c. 1
x 2 3y 2 2z 5 29
2
2x 2 5y 1 z 5 3
3 23x 1 6y 1 2z 5 8
Equation 1 1 (2 3 equation 2 ): 5x 2 13y 5 23
Equation 2 1 ( equation 3 ): 22x 1 3y 5 21
2(5x 2 13y 5 23)
1 5(22x 1 3y 5 21)
211y 5 211
y51
5x 2 13(1) 5 23
x52
(2) 2 3(1) 2 2z 5 29
z54
(2, 1, 4)
13. a. The two lines intersect at a point.
b. The two planes are parallel and do not meet.
c. The three planes intersect at a point.
14. a. L:(x 2 y 5 1) 1 (y 1 z 5 23)
5 x 1 z 5 22
1
L1: y 2 z 5 0, x 5 2
2
x 1 z 5 22
1
a2 b 1 z 5 22
2
3
z52
2
y2z50
Chapter 9: Relationships Between Points, Lines, and Planes
3
y 2 a2 b 5 0
2
y52
3
2
1 3 3
a2 , 2 , 2 b
2 2 2
0 n ? n1 0
b. cos u 5
0 n 0 0 n1 0
n 5 (1, 1, 21)
n1 5 (0, 1, 1)
0
cos u 5
@ "3 @ @ "2 @
u 5 90 °
c. (0, 1, 1) 3 (1, 1, 21) 5 (22, 1, 21)
5 (2, 21, 1)
Ax 1 By 1 Cz 1 D 5 0
2x 2 y 1 z 1 D 5 0
21
23
23
2a
b2a
b1a
b1D50
2
2
2
D51
2x 2 y 1 z 1 1 5 0
9.4 The Intersection of Three Planes,
pp. 531–533
x 2 3y 1 z 5 2
0x 1 y 2 z 5 21
3 0x 1 0y 1 3z 5 212
The system can be solved by first solving equation 3
for z. Thus,
3z 5 212
z 5 24
If we use the method of back substitution, we can
substitute z 5 24 into equation 2 and solve for y.
y 2 (24) 5 21
y 5 25
If we substitute y 5 25 and z 5 24 into equation 1
we obtain the value of x.
x 2 3(25) 2 4 5 2 or x 5 29
The three planes intersect at the point with
coordinates (29, 25, 24)
Check:
Substituting into equation 1 :
x 2 3y 1 z 5 29 1 15 2 4 5 2
Substituting into equation 2 :
0x 1 y 2 z 5 25 1 4 5 21
Substituting into equation 3 : 0x 1 0y 1 3z 5 212
1. a.
1
2
Calculus and Vectors Solutions Manual
b. This solution is the point at which all three
planes meet.
2. a. 1
x2y1z54
2 0x 1 0y 1 0z 5 0
3 0x 1 0y 1 0z 5 0
The answer may vary depending upon the constant
you multiply the equations by. For example,
2 3 (x 2 y 1 z 5 4) 5 2x 2 2y 1 2z 5 8
3 3 (x 2 y 1 z 5 4) 5 3x 2 3y 1 3z 5 12
3x 2 3y 1 3z 5 12 and 2x 2 2y 1 2z 5 8 are
equations that could work.
b. These three planes are intersecting in one single
plane, because all three equations can be changed
into one equivalent equation. They are coincident
planes.
c. Setting x 5 t and y 5 s leads to
t 2 s 1 z 5 4 or z 5 s 2 t 1 4, s, tPR
d. Setting y 5 t and z 5 s leads to
x 2 t 1 s 5 4 or x 5 t 2 s 1 4, s, tPR
3. a. 1 2x 2 y 1 3z 5 22
2
x 2 y 1 4z 5 3
3 0x 1 0y 1 0z 5 1
The answer may vary depending upon the constants
and equations you use to determine your answer.
For example,
Equation 1 1 equation 2 1 equation 3 5
(2x 2 y 1 3z 5 22)
1 (x 2 y 1 4z 5 3)
1 (0x 1 0y 1 0z 5 1)
3x 2 2y 1 7z 5 2
or
2 3 equation 2 2 equation 3 5
(2x 2 2y 1 8z 5 6)
2 (0x 1 0y 1 0z 5 1)
2x 2 2y 1 8z 5 5
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
3x 2 2y 1 7z 5 2 is one system of equations that
could produce the original system composed of
equations 1 , 2 , and 3 .
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
2x 2 2y 1 8z 5 5 is another system of equations
that could produce the original system composed of
equations 1 , 2 , and 3 .
b. The systems have no solutions.
4. a. 1
x 1 2y 2 z 5 4
2 x 1 0y 2 2z 5 0
3 2x 1 0y 1 0z 5 26
The system can be solved by first solving equation 3
for x. So,
9-15
2x 5 26
x 5 23
If we use the method of back substitution, we can
substitute x 5 23 into equation 2 and solve for z.
23 2 2z 5 0
3
z52
2
If we substitute x 5 23 and z 5 2 32 into equation 1
we obtain the value of y.
11
3
23 1 2y 1 5 4 or y 5
2
4
The equations intersect at the point with coordinates
(23, 114, 2 32 )
Check:
Substituting into equation 1 :
x 1 2y 2 z 5 23 1 224 1 32 5 4
Substituting into equation 2 :
x 1 0y 2 2z 5 23 1 3 5 0
Substituting into equation 3 : 2x 1 0y 1 0z 5 26
b. This solution is the point at which all three
planes meet.
5. a. 1
2x 2 y 1 z 5 1
2
x 1 y 2 z 5 21
3 23x 2 3y 1 3z 5 3
Since equation 3 5 2 equation 2 , equation 2
and equation 3 are consistent or lie in the same
plane. Equation 1 meets this plane in a line.
b. Adding equation 2 and equation 1 creates an
equivalent equation, 3x 5 0 or x 5 0. Substituting
x 5 0 into equation 1 and equation 2 gives
equation 4 z 2 y 5 1 and equation 5
y 2 z 5 21. Equations 4 and 5 indicate the
problem has infinite solutions. Substituting y 5 t
into equation 4 or 5 leads to
x 5 0, y 5 t, and z 5 1 1 t, tPR
Check:
2(0) 2 s 1 (s 1 1) 5 1
0 1 s 2 (s 1 1) 5 21
23(0) 2 3(s) 1 3(s 1 1) 5 3
6. 1 2x 1 3y 2 4z 5 25
2
x 2 y 1 3z 5 2201
3 5x 2 5y 1 15z 5 21004
There is no solution to this system of equations,
because if you multiply equation 2 by 5 you
obtain a new equation, 5x 2 5y 1 15z 5 21005,
which is inconsistent with equation 3 .
7. a. Yes when this equation is alone, this is true,
because any constants can be substituted into the
variables in the equation 0x 1 0y 1 0z 5 0 and the
equation will always be consistent.
9-16
b. Answers may vary. For example: To obtain a no
solution and an equation with 0x 1 0y 1 0z 5 0,
you must have two equal planes and one parallel
distinct plane. For example one solution is:
x1y1z52
2x 1 2y 1 2z 5 4
3x 1 3y 1 3z 5 12
8. a. 1 2x 1 y 2 z 5 23
2 x 2 y 1 2z 5 0
3 3x 1 2y 2 z 5 25
2 3 equation 2 1 equation 3 5 5x 1 0y 1 0z
5 25 which gives x 5 21.
Equation 1 1 equation 2 5 3x 1 0y 1 1z
5 23. Substituting x 5 1 into this equation leads
to: 3(21) 1 z 5 23 or z 5 0.
Substituting z 5 0 and x 5 21 into equation 1
gives: 2(21)y 2 0 5 23 or y 5 21. (21, 21, 0)
is the point at which the three planes meet.
Check:
Substituting into equation 1 :
2x 1 y 2 z 5 22 2 1 1 0 5 23
Substituting into equation 2 :
x 2 y 1 2z 5 21 1 1 1 0 5 0
Substituting into equation 3 :
3x 1 2y 2 z 5 23 2 2 1 0 5 25
x
y
7
b. 1
2 1z5
3
4
8
2 2x 1 2y 2 3z 5 220
3
x 2 2y 1 3z 5 2
Equation 2 1 equation 3 5 3x 1 0y 1 0z 5 218
which gives x 5 26.
Equation 3 2 3 3 Equation 1 5 2 54y 5 2 58 or
y 5 12. Substituting x 5 26 and y 5 12 into equation 3
leads to:
1
26 2 2a b 1 3z 5 2 or z 5 3.
2
(26, 12, 3) is the point at which the three planes meet.
Check:
Substituting into equation 1 :
x
y
1
7
2 4 1 z 5 22 2 8 1 3 5 8
3
Substituting into equation 2 :
2x 1 2y 2 3z 5 212 1 1 2 9 5 220
Substituting into equation 3 :
x 2 2y 1 3z 5 26 2 1 1 9 5 2
c. 1 x 2 y 5 2199
2 x 1 z 5 2200
3 y 2 z 5 201
Equation 2 1 equation 3 5 equation 4
5x1y51
Chapter 9: Relationships Between Points, Lines, and Planes
Equation 4 1 equation 1 5 2x 5 2198 or
x 5 299. Substituting x 5 299 into equation 1
leads to:
299 2 y 5 2199 or y 5 100. Substituting x 5 299
into equation 2 , you obtain:
299 1 z 5 2200 or z 5 2101
(299, 100, 2101) is the point at which the three
planes meet.
Check:
Substituting into equation 1 :
x 2 y 5 299 2 (100) 5 2199
Substituting into equation 2 :
x 1 z 5 299 2 101 5 2200
Substituting into equation 3 :
y 2 z 5 100 2 (2101) 5 201
d. 1 x 2 y 2 z 5 21
2
y2250
3
x1155
Rearranging equation 2 gives y 5 2. Solving for x
in equation 3 gives x 5 4.
Substituting x 5 4 and y 5 2 into equation 1
leads to:
4 2 2 2 z 5 21 or z 5 3.
(4, 2, 3) is the point at which all three planes meet.
9. a. 1 x 2 2y 1 z 5 3
2 2x 1 3y 2 z 5 29
3 5x 2 3y 1 2z 5 0
Equation 3 1 equation 2 5 equation 4
5 7x 1 1z 5 29.
Setting z 5 t, x 5 2 17 t 2 97
Equation 2 2 2 3 equation 1 5 equation 5
5 7y 1 23z 5 215.
Setting z 5 t, y 5 2 157 1 37 t
x 5 2 17 t 2 97, y 5 2 157 1 37 t, and z 5 t, tPR The
planes intersect in a line.
b. 1 x 2 2y 1 z 5 3
2
x1y1z52
3 x 2 3y 1 z 5 26
Equation 3 2 equation 2 5 24y 5 28 or y 5 2
Equation 3 2 equation 1 5 21y 5 29 or y 5 9
Since the solutions for y are different from these
two equations, there is no solution to this system of
equations.
c. 1 x 2 y 1 z 5 22
2 x1y1z52
3 x 2 3y 1 z 5 26
Equation 1 1 equation 2 5 equation 4
5 2x 1 2y 5 0.
Setting z 5 t, x 5 2t
Calculus and Vectors Solutions Manual
Using z 5 t and x 5 2t, Solve equation 1
2t 2 y 1 t 5 22 or y 5 2
x 5 2t, y 5 2, and z 5 t, tPR
The planes intersect in a line.
10. a. 1
x2y1z52
2 2x 2 2y 1 2z 5 4
3
x 1 y 2 z 5 22
Equation 1 1 equation 3 5 equation 4
5 2x 5 0 or x 5 0.
Setting z 5 t,
Equation 1 : 0 2 y 1 t 5 2 or y 5 t 2 2
x 5 0, y 5 t 2 2, and z 5 t, tPR
b. 1 2x 2 y 1 3z 5 0
2 4x 2 2y 1 6z 5 0
3 22x 1 y 2 3z 5 0
Equation 1 1 equation 3 5 equation 4
5 2x 5 0 or x 5 0.
Setting y 5 t and z 5 s, equation 1 :
t 2 3s
2x 2 t 1 3s 5 0 or x 5
2
t 2 3s
x5
, y 5 t, and z 5 s, s, tPR
2
x1y1z51
x 2 2y 1 z 5 0
3
x2y1z50
1
Equation
2 equation 3 5 equation 4
5 2y 5 1 or y 5 12
Equation 2 2 equation 3 5 equation 5
5 2y 5 0 or y 5 0
Since the y-variable is different in equation 4 and
equation 5 , the system is inconsistent and has no
solution.
b. Answers may vary. For example: If you use the
normals from equations 1 , 2 , and 3 , you can
determine the direction vectors from the equations’
coefficients.
>
n1 5 (1, 1, 1)
>
n2 5 (1, 22, 1)
>
n3 5 (1, 21, 1)
>
>
m1 5 n1 3 n2 5 (3, 0, 23)
>
>
m2 5 n1 3 n3 5 (2, 0, 22)
>
>
m3 5 n2 3 n3 5 (21, 0, 1)
c. The three lines of intersection are parallel and are
pairwise coplanar, so they form a triangular prism.
>
>
>
d. n1 3 n2 is perpendicular to n3 . So since,
>
>
>
(n1 3 n2 ) ? n3 5 0, a triangular prism forms.
12. a. 1 x 2 y 1 3z 5 3
2 x 2 y 1 3z 5 6
3
3x 2 5z 5 0
11. a.
1
2
9-17
Equation 1 and equation 2 have the same set of
coefficients and variables, however, equations 1
equals 3 while equation 2 equals 6, which means
there is no possible solution.
b. 1 5x 2 2y 1 3z 5 1
2 5x 2 2y 1 3z 5 21
3 5x 2 2y 1 3z 5 13
All three equations equal different numbers so there
is no possible solution.
c. 1
x2y1z59
2 2x 2 2y 1 2z 5 18
3 2x 2 2y 1 2z 5 17
Equation 2 equals 18 while equation 3 equals 17,
which means there is no possible solution.
d. The coefficients of equation 1 are half the
coefficients of equation 2 , but the constant term
is not half the other constant term.
13. a. 1 2x 2 y 2 z 5 10
2 x 1 y 1 0z 5 7
3 0x 1 y 2 z 5 8
Equation 1 2 2 3 equation 2 2 equation 3 :
24y 5 2 12 or y 5 3. Substituting y 5 3 into
equation 2 and equation 3 gives:
x 1 3 1 0z 5 7 or x 5 4
0x 1 3 2 z 5 8 or z 5 25
(4, 3, 25)
b. 1 2x 2 y 1 z 5 23
2
x 1 y 2 2z 5 1
3 5x 1 2y 2 5z 5 0
Equation 1 1 equation 2 : 3x 2 z 5 22.
Setting z 5 t, x 5
Equation
Setting
1
t22
3
2 2 3 equation
2
: 23y 1 5z 5 25.
5t 1 5
z 5 t, y 5 3
t22
5t 1 5
,y5
, z 5 t, tPR
3
3
c. 1
x1y2z50
2
2x 2 y 1 z 5 0
3 4x 2 5y 1 5z 5 0
Equation 1 1 equation 2 : 3x 5 0 or x 5 0
Setting x 5 0 and z 5 t in equation 2 gives,
2(0) 2 y 1 t 5 0 or y 5 t
x 5 0, y 5 t, z 5 t, tPR
d. 1 x 2 10y 1 13z 5 24
2 2x 2 20y 1 26z 5 28
3 x 2 10y 1 13z 5 28
If you multiply equation 2 by two, you obtain
2x 2 20y 1 26z 5 216. Since equation 2 and
x5
9-18
equation 3 equal different numbers, there is no
solution to this system.
e. 1 x 2 y 1 z 5 22
2
x1y1z52
3 3x 1 y 1 3z 5 2
Equation 1 1 equation 2 : 22y 5 24 or y 5 2
Setting y 5 2 and z 5 t in equation 1 ,
x 2 2 1 t 5 22 or x 5 2t
x 5 2t, y 5 2, z 5 t, tPR
f. 1 x 1 y 1 z 5 0
2 x 2 2y 1 3z 5 0
3 2x 2 y 1 3z 5 0
Equation 1 2 equation 2 5 equation 4
5 3y 2 2z 5 0
Equation 3 2 2 3 equation 2 2 equation 5
5 3y 2 3z 5 0
Equation 4 2 equation 5 : z 5 0
Setting z 5 0 in equation 1 and equation 2 ,
Equation 6 5 x 1 y 5 0
Equation 7 5 x 2 2y 5 0
Equation 6 2 equation 7 : 3y 5 0 or y 5 0
Setting y 5 0 and z 5 0 in equation 1 leads to
x50
(0, 0, 0)
14. a. First, reorder these equations so that equation
2 is first, equation 3 is second, and equation 1
last.
1
x2y1z5p
2 4x 1 qy 1 z 5 2
3 2x 1 y 1 z 5 4
To eliminate x from the last two equations, subtract
4 times equation 1 from equation 2 , and subtract
2 times equation 1 from equation 3 .
1
x2y1z5p
2 (q 1 4)y 2 3z 5 2 2 4p
3
3y 2 z 5 4 2 2p
There will be an infinite number of solutions if
q 1 4 5 9 and 3(4 2 2p) 5 2 2 4p because then
equation 2 will be 3 times equation 3 . This means
that p 5 q 5 5.
b. Based on what was found in part a., substituting
in p 5 q 5 5 we will arrive at the equivalent system
1 x2y1z55
2
9y 2 3z 5 218
3
3y 2 z 5 26
which is really the same as
1 x2y1z55
2
3y 2 z 5 26
Letting z 5 t, we see that equation 2 delivers
Chapter 9: Relationships Between Points, Lines, and Planes
1
(t 2 6)
3
1
5 t22
3
and so equation 1 gives
1
x 5 (t 2 6) 2 t 1 5
3
2
52 t13
3
So the parametric equation of the line of
intersection is
2
1
x 5 2 t 1 3, y 5 t 2 2, z 5 t, tPR.
3
3
15. a. First, eliminate x from two of these equations.
To make things easier, switch equation 1 with
equation 2 , and multiply equation 3 by 2.
1
2x 1 y 1 z 5 24
2
4x 1 3y 1 3z 5 28
3 6x 2 4y 1 (2m 2 2 12)z 5 2m 2 8
Now eliminate x from the last two equations by
using proper multiples of the first equation.
1
2x 1 y 1 z 5 24
2
y1z50
3 27y 1 (2m 2 2 15)z 5 2m 1 4
Now eliminate y from the third equation by using a
proper multiple of the second equation.
1 2x 1 y 1 z 5 24
2
y1z50
3 (2m 2 2 8)z 5 2m 1 4
If 2m 2 2 8 5 0 (the coefficient of z in the third
equation), then m 5 6 2. However, if m 5 2, the
third equation would become 0z 5 8, which has no
solutions. So there is no solution if m 5 2.
b. Working with what was found in part a., if m 2 62,
then the third equation in the equivalent system found
there will have a unique solution for z, namely
2m 1 4
z5
,
2m 2 2 8
and back-substituting into the other two equations
will give unique solutions for x and y also. So there
is a unique solution if m 2 62.
c. Again using the equivalent system found in part a.,
setting m 5 22 will deliver the third equation 0z 5 0,
which allows for z to be anything at all. So m 5 22
will give an infinite number of solutions.
1
1
1
16. a. 1
1 2 50
a
c
b
2
2
3
13
2
1 1 5
a
c
b
6
y5
Calculus and Vectors Solutions Manual
4
3
2
5
2 1 5
a
c
b
2
Equation 2 2 2 3 equation
1
4
13
1 5
5 equation 4
c
b
6
Equation 3 2 4 3 equation
3
1
:
1
6
7
:2 1
c
b
5
>
>
m3 5 n 3 n1 5 (21, 0, 1) 5 5 equation
2
Equation 5 1 6 3 equation 4 :
31
5 15.5 or c 5 2
c
Substituting c 5 2 into equation 4 :
13
1
or b 5 6
125
b
6
Substituting c 5 2 and b 5 6 into equation
1
1
1
1 2 5 0 or a 5 3
a
6
2
(3, 6, 2)
5
1
:
9.5 The Distance from a Point to a
Line in R2 and R3, pp. 540–541
1. a. 3x 1 4y 2 5 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(24) 1 4(5) 2 5 0
d5
"32 1 42
3
5
5
b. 5x 2 12y 1 24 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 5(24) 2 12(5) 1 24 0
d5
"52 1 (212)2
56
or 4.31
5
13
c. 9x 2 40y 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 9(24) 2 40(5) 0
d5
"92 1 (40)2
236
5
or 5.76
"1681
2. a. 2x 2 y 1 1 5 0 and 2x 2 y 1 6 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
9-19
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
2(0) 2 y 1 1 5 0 or y 5 1 which corresponds to
the point (0, 1).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 2(0) 2 1(1) 1 6 0
d5
"22 1 (21)2
5
or 2.24
5
"5
b. 7x 2 24y 1 168 5 0 and 7x 2 24y 2 336 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
7(0) 2 24y 1 168 5 0 or y 5 7 which corresponds
to the point (0, 7)
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 7(0) 2 24(7) 2 336 0
d5
"72 1 (224)2
504
or 20.16
5
25
>
3. a. r 5 (21, 2) 1 s(3, 4), sPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 21 1 3s,
y 5 2 1 4s. We construct a vector from R(22, 3)
to a general point on the line.
>
a 5 322 2 (21 1 3s), 3 2 (2 1 4s)4
5 (21 2 3s, 1 2 4s).
(3, 4) ? (21 2 3s, 1 2 4s) 5 0
(23 2 9s) 1 (4 2 16s) 5 0
1
s5
25
This means that the minimal distance between
R(22, 3) and the line occurs when s 5 251 .
54
This point corresponds to A2 22
25 , 25 B. The distance
between this point and (22, 3) is 1.4.
>
b. r 5 (1, 0) 1 t(5, 12), tPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 5t,
y 5 12t. We construct a vector from R(22, 3) to a
general point on the line.
>
a 5 322 2 (1 1 5t), 3 2 (12t)4
5 (23 2 5t, 3 2 12t).
(5, 12) ? (23 2 5t, 3 2 12t) 5 0
9-20
(215 2 25t) 1 (36 2 144t) 5 0
21
t5
169
This means that the minimal distance between
21
R(22, 3) and the line occurs when t 5 169
.
274 252
This point corresponds to A 169, 169 B. The distance
between this point and (22, 3) is about 3.92.
>
c. r 5 (1, 3) 1 p(7, 224), pPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 7p,
y 5 3 2 24p. We construct a vector from R(22, 3)
to a general point on the line.
>
a 5 322 2 (1 1 7p), 3 2 (3 2 24p)4
5 (23 2 7p, 24p).
(7, 224) ? (23 2 7p, 24p) 5 0
(221 2 49p) 1 (2576p) 5 0
21
p52
625
This means that the minimal distance between
21
R(22, 3) and the line occurs when p 5 2 625
.
478 2379
This point corresponds to A 625, 625 B.
The distance between this point and (22, 3) is
about 2.88.
0 Ax0 1 By0 1 C 0
4. a. d 5
"A 2 1 B 2
If you substitute in the coordinates (0, 0), the
0 A(0) 1 B(0) 1 C 0
formula changes to d 5
,
"A 2 1 B 2
0C0
which reduces to d 5
.
"A 2 1 B 2
b. 3x 2 4y 2 12 5 0 and 3x 2 4y 1 12 5 0
0C0
0 212 0
d(L1 ) 5
5
"A 2 1 B 2
"32 1 (24)2
12
5
5
0C0
0 12 0
d(L2 ) 5
5
2
2
2
"A 1 B
"3 1 (24)2
12
5
5
The distance between these parallel lines is
12
12
24
5 1 5 5 5 , because one of the lines is below
the origin and the other is above the origin.
c. 3x 2 4y 2 12 5 0 and 3x 2 4y 1 12 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
Chapter 9: Relationships Between Points, Lines, and Planes
3(0) 2 4y 2 12 5 0 or y 5 23 which corresponds
to the point (0, 3).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(0) 2 4(23) 1 12 0
d5
"32 1 (24)2
24
5
5
Both the answers to 4.b. and 4.c. are the same.
>
5. a. r 5 (22, 1) 1 s(3, 4), sPR
>
r 5 (1, 0) 1 t(3, 4), tPR
First find a random point on one of the lines. We
will use (22, 1) from the first equation. We start by
writing the second equation in parametric form.
Doing so gives x 5 1 1 3t, y 5 4t. We construct a
vector from P(22, 1) to a general point on the line.
>
a 5 322 2 (1 1 3t), 1 2 (4t)4
5 (23 2 3t, 1 2 4t).
(3, 4) ? (23 2 3t, 1 2 4t) 5 0
(29 2 9t) 1 (4 2 16t) 5 0
1
t52
5
This means that the minimal distance between
P(22, 1) and line occurs when t 5 2 15. This point
corresponds to A 25, 2 45 B. The distance between this
point and (22, 1) is 3
x
x21
y
y11
and 5
5
4
23
4
23
First change one equation into a Cartesian equation,
which leads to 3x 1 4y 2 3 5 0 and take a point
from the other equation such as (4, 24).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(4) 1 4(24) 2 3 0
d5
"32 1 42
7
5 or 1.4
5
b.
c. 2x 2 3y 1 1 5 0 and 2x 2 3y 2 3 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
2(0) 2 3y 2 3 5 0 or y 5 21 which corresponds
to the point (0, 21).
Calculus and Vectors Solutions Manual
d5
d5
5
0 Ax0 1 By0 1 C 0
"A 2 1 B 2
0 2(0) 2 3(21) 1 1 0
4
"22 1 (23)2
or 1.11
"13
d. 5x 1 12y 5 120 and 5x 1 12y 1 120 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
5(0) 1 12y 5 120 or y 5 10 which corresponds to
the point (0, 10).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 5(0) 1 12(10) 1 120 0
d5
"52 1 122
240
or 18.46
5
13
>
6. a. P(1, 2, 21) r 5 (1, 0, 0) 1 s(2, 21, 2), sPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 2s,
y 5 2s, and z 5 2s. We construct a vector from
P(1, 2, 21) to a general point on the line.
>
a 5 31 2 (1 1 2s), 2 2 (2s), 21 2 (2s)4
5 (22s, 2 1 s, 21 2 2s).
(2, 21, 2) ? (22s, 2 1 s, 21 2 2s) 5 0
(24s) 1 (22 2 s) 1 (22 2 4s) 5 0
4
s52
9
This means that the minimal distance between
P(1, 2, 21) and the line occurs when s 5 2 49.
This point corresponds to A 19, 49, 2 89 B. The distance
between this point and P(1, 2, 21) is 1.80.
>
b. P(0, 21, 0) r 5 (2, 1, 0) 1 t(24, 5, 20), tPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 2 2 4t,
y 5 1 1 5t, and z 5 20t. We construct a vector
from P(0, 21, 0) to a general point on the line.
>
a 5 30 2 (2 2 4t), 21 2 (1 1 5t), 0 2 (20t)4
5 (22 1 4t, 22 2 5t, 20t).
(24, 5, 20) ? (22 1 4t, 22 2 5t, 220t) 5 0
(8 2 16t) 1 (210 2 25t) 1 (2400t) 5 0
2
t52
441
This means that the minimal distance between
2
P(0, 21, 0) and the line occurs when t 5 2 441
.
9-21
431
40
This point corresponds to A 890
441 , 441 , 2 441 B. The
distance between this point and P(0, 21, 0) is 2.83.
>
c. P(2, 3, 1) r 5 p(12, 23, 4), pPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 12p, y 5 23p,
and z 5 4p. We construct a vector from P(2, 3, 1)
to a general point on the line.
>
a 5 32 2 (12p), 3 2 (23p), 1 2 (4p)4
5 (2 2 12p, 3 1 3p, 1 2 4p).
(12, 23, 4) ? (2 2 12p, 3 1 3p, 1 2 4p) 5 0
(24 2 144p) 1 (29 2 9p) 1 (4 2 16p) 5 0
19
p5
169
This means that the minimal distance between
19
P(2, 3, 1) and the line occurs when p 5 169
. This
57 76
point corresponds to A 228
169 , 2 169 , 169 B. The distance
between this point and P(2, 3, 1) is 3.44.
>
7. a. r 5 (1, 1, 0) 1 s(2, 1, 2), sPR
>
r 5 (21, 1, 2) 1 t(2, 1, 2), tPR
First find a random point on one of the lines. We
will use P(21, 1, 2) from the second equation. We
then write the first equation in parametric form.
Doing so gives x 5 1 1 2s, y 5 1 1 s, and
z 5 0 1 2s. We construct a vector from P(21, 1, 2)
to a general point on the line.
>
a 5 321 2 (1 1 2s), 1 2 (1 1 s), 2 2 2s4
5 (22 2 2s, 2 2 2s).
(2, 1, 2) ? (22 2 2s, 2s, 2 2 2s) 5 0
(24 2 4s) 1 (2s) 1 (4 2 4s) 5 0
s50
This means that the minimal distance between
P(21, 1, 2) and line occurs when s 5 0. This point
corresponds to (1, 1, 0). The distance between this
point and (21, 1, 2) is 2.83
>
b. r 5 (3, 1, 22) 1 m(1, 1, 3), mPR
>
r 5 (1, 0, 1) 1 n(1, 1, 3), nPR
First find a random point on one of the lines.
We will use P(1, 0, 1) from the second equation.
We then write the first equation in parametric form.
Doing so gives x 5 3 1 m, y 5 1 1 m, and
z 5 22 1 3m. We construct a vector from P(1, 0, 1)
to a general point on the line.
>
a 5 31 2 (3 1 m), 0 2 (1 1 m), 1 2 (22 1 3m)4
5 (22 2 3m, 21 2 m, 3 2 3m).
(1, 1, 3) ? (22 2 3m, 21 2 m, 3 2 3m) 5 0
(22 2 3m) 1 (21 2 m) 1 (9 2 9m) 5 0
6
m5
13
9-22
This means that the minimal distance between
P(1, 0, 1) and line occurs when m 5 136 . This point
19
6
corresponds to A 45
13 , 13 , 2 13 B. The distance between
this point and (1, 0, 1) is 3.28
>
8. a. r 5 (1, 21, 2) 1 s(1, 3, 21), sPR
First we write the equation in parametric form.
Doing so gives x 5 1 1 s, y 5 21 1 3s, and
z 5 2 2 s. We construct a vector from P(2, 1, 3) to
a general point on the line.
>
a 5 32 2 (1 1 s), 1 2 (21 1 3s), 3 2 (2 2 s)4
5 (1 2 s, 2 2 3s, 1 1 s).
(1, 3, 21) ? (1 2 s, 2 2 3s, 1 1 s) 5 0
(1 2 s) 1 (6 2 9s) 1 (1 1 s) 5 0
6
s5
11
This means that the minimal distance between
P(2, 1, 3) and line occurs when s 5 116 . This point
7 16
corresponds to A 17
11 , 11 , 11 B.
7 16
b. The distance between A 17
11 , 11 , 11 B and (2, 1, 3)
is 1.65.
9. First, find the line L of intersection between the
planes
1 x 2 y 1 2z 5 2
2
x 1 y 2 z 5 22
Subtract the first equation from the second to
eliminate x and get the equivalent system
1 x 2 y 1 2z 5 2
2
2y 2 3z 5 24
Let z 5 t. Then the second equation gives
2y 5 3t 2 4
3
y5 t22
2
So substituting these into the first equation gives
x 5 y 2 2z 1 2
3
5 a t 2 2b 2 2t 1 2
2
1
52 t
2
So the equation of the line of intersection for these
two planes in parametric form is
1
3
x 5 2 t, y 5 t 2 2, z 5 t, tPR.
2
2
The direction vector for this line is A2 12, 32, 1B,
which is parallel to (21, 3, 2). So, to make things
easier, the parametric form of this line of
intersection could also be expressed as
x 5 2t, y 5 3t 2 2, z 5 2t, tPR
In vector form, this is the same as
Chapter 9: Relationships Between Points, Lines, and Planes
>
r 5 (0, 22, 0) 1 t(21, 3, 2), tPR.
Since Q(0, 22, 0) is on this line,
>
QP 5 (21, 2, 21) 2 (0, 22, 0)
5 (21, 4, 21)
So the distance from P(21, 2, 21) to the line of
intersection is
0 (21, 3, 2) 3 (21, 4, 21) 0
d5
0 (21, 3, 2) 0
0 (211, 23, 21) 0
5
0 (21, 3, 2) 0
131
Å 14
8 3.06
To find the point on the line that gives this minimal
distance, let (x, y, z) be a point on the line. Then,
using the parametric equations,
(x, y, z) 5 (2t, 3t 2 2, 2t)
So the distance from P to this point is
"(x 1 1)2 1 (y 2 2)2 1 (z 1 1)2
5 "(1 2 t)2 1 (3t 2 4)2 1 (2t 1 1)2
5
5 "14t 2 2 22t 1 18
To get the minimal distance, set this quantity
equal to #131
14 .
"14t 2 2 22t 1 18 5
131
Å 14
131
14t 2 2 22t 1 18 5
14
196t 2 2 308t 1 252 5 131
196t 2 2 308t 1 121 5 0
308 6 "0
t5
392
11
5
14
So the point on the line at minimal distance from P is
(x, y, z) 5 (2t, 3t, 22, 2t)
11
11
11
5 a2 , 3a b 2 2, 2a bb
14
14
14
11 5 22
5 a2 , , b
14 14 14
10. A point on the line
>
r 5 (0, 0, 1) 1 s(4, 2, 1), sPR.
has parametric equations
x 5 4s, y 5 2s, z 5 1 1 s, sPR.
Let this point be called
Q(4s, 2s, 1 1 s). Then
>
QA 5 (2, 4, 25) 2 (4s, 2s, 1 1 s)
5 (2 2 4s, 4 2 2s, 26 2 s)
Calculus and Vectors Solutions Manual
If Q is at minimal distance from A, then this vector
will be perpendicular to the direction vector for the
line, (4, 2, 1). This means that
0 5 (2 2 4s, 4 2 2s, 26 2 s) ? (4, 2, 1)
5 10 2 21s
10
s5
21
So the point Q on the line at minimal distance from
A is
10
10
10
Q(4s, 2s, 1 1 s) 5 Qa4a b, 2a b, 1 1 b
21
21
21
40 20 31
5 Qa , , b
21 21 21
Also
>
40
20
31
QA 5 a2 2 , 4 2 , 25 2 b
21
21
21
2 64 136
5 a , ,2
b
21 21
21
So the point Ar will satisfy
>
>
QAr 5 2QA
2
64 136
5 a2 , 2 ,
b
21 21 21
5 Ar(a, b, c) 2 Q
40
20
31
5 aa 2 , b 2 , c 2 b
21
21
21
38
44
167
So a 5 21, b 5 2 21, and c 5 21 . That is,
44 167
Ar( 38
21 , 2 21 , 21 ).
11. a. Think of H as being the origin, E as being on
the x-axis, D as being on the z-axis, and G as being
on the y-axis. That is,
H(0, 0, 0)
E(3, 0, 0)
G(0, 2, 0)
D(0, 0, 2)
and so on for the other points as well. Then line
segment HB has direction vector
B(3, 2, 2) 2 H(0, 0, 0) 5 (3, 2, 2).
>
Also, HA 5 (3, 0, 2). So the distance formula says
that the distance between A and line segment HB is
0 (3, 2, 2) 3 (3, 0, 2) 0
d5
0 (3, 2, 2) 0
0 4, 0, 26 0
5
0 (3, 2, 2) 0
52
Å 17
8 1.75
5
9-23
b. Vertices D and G will give the same distance to
HB because they are equidistant to the segment HB.
(This is easy to check with the distance formula
used similarly to part a. The vertices C, E, and F
give different distances than those found in part a.)
c. The height of triangle AHB was found in part a.,
and was # 52
base length of this triangle is the
17 . The
>
magnitude of HB 5 (3, 2, 2), which is # 52. So
the area of this triangle is
1
52
1
a
b ("17) 5 ("52)
2 Å 17
2
8 3.6 units 2
9.6 The Distance from a Point to a
Plane, pp. 549–550
1. a. Yes the calculations are correct, Point A lies in
the plane.
b. The answer 0 means that the point lies in the
plane.
2. Use the distance formula.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. The distance from A (3, 1, 0) to the plane
20x 2 4y 1 5z 1 7 5 0 is
0 20(3) 1 24(1) 1 5(0) 1 7 0
d5
"202 1 (24)2 1 52
53
b. The distance from B(0, 21, 0) to the plane
2x 1 y 1 2z 2 8 5 0 is
0 2(0) 1 1(21) 1 2(0) 2 8 0
d5
"22 1 12 1 22
53
c. The distance from C(5, 1, 4) to the plane
3x 2 4y 2 1 5 0 is
0 3(5) 1 24(1) 1 0(4) 2 1 0
d5
"32 1 (24)2 1 02
52
d. The distance from D(1, 0, 0) to the plane
5x 2 12y 5 0 is
0 5(1) 2 12(0) 1 0(0) 1 0 0
d5
"5 2 1 (212)2 1 02
5
or 0.38
5
13
9-24
e. The distance from E(21, 0, 1) to the plane
18x 2 9y 1 18z 2 11 5 0 is
0 18(21) 2 9(0) 1 18(1) 2 11 0
d5
"182 1 (29)2 1 182
11
or 0.41
5
27
3. a. 3x 1 4y 2 12z 2 26 5 0 and
3x 1 4y 2 12z 1 39 5 0
First find a point in the second plane such as
(23, 0, 0). Then use d 5
0 Ax0 1 By0 1 Cz0 1 D 0
!A2 1 B2 1 C2
to solve.
0 3(213) 1 4(0) 2 12(0) 2 26 0
d5
"32 1 42 1 (212)2
55
b. 3x 1 4y 2 12z 2 26 5 0
1 3x 1 4y 2 12z 1 39 5 0
6x 1 8y 2 24z 1 13 5 0
c. Answers may vary. Any point on the plane
6x 1 8y 2 24z 1 13 5 0 will work, for example
(2 16, 0, 12) .
4. a. The distance from P(1, 1, 23) to the plane
y 1 3 5 0 is
0 0(1) 1 1(1) 1 0(23) 1 3 0
d5
"02 1 (1)2 1 02
54
b. The distance from Q(21, 1, 4) to the plane
x 2 3 5 0 is
0 1(21) 1 0(1) 1 0(4) 2 3 0
d5
"12 1 02 1 02
54
c. The distance from R(1, 0, 1) to the plane
z 1 1 5 0 is
0 0(1) 1 0(0) 1 1(1) 1 1 0
d5
"02 1 02 1 12
52
5. First you have to find an equation of a plane to
the three points. The equation to this plane is
14x 2 28y 1 28z 2 42 5 0. Then use
d5
0 Ax0 1 By0 1 Cz0 1 D 0
d5
0 14(1) 2 28(21) 1 28(1) 2 42 0
5
!A2 1 B2 1 C2
to solve for the distance.
"142 1 (228)2 1 282
2
or 0.67
3
Chapter 9: Relationships Between Points, Lines, and Planes
6. 3 5
0 A(3) 2 2(23) 1 6(1) 1 0 0
"A 1 (22) 1 6
3" ( A 1 40 ) 5 Z 3A 1 12 Z
" ( A2 1 40 ) 5 Z A 1 4 Z
A2 1 40 5 A2 1 8A 1 16
24 5 8A
35A
A 5 3 is the only solution to this equation.
7. These lines are skew lines, and the plane containing
>
the second line, r 5 (0, 0, 1) 1 t(1, 1, 0), tPR,
that is parallel to the first line will have direction
vectors (1, 1, 0) and (3, 0, 1). So a normal to this
plane is (1, 1, 0) 3 (3, 0, 1) 5 (1, 21, 23).
So the equation of this plane will be of the form
x 2 y 2 3z 1 D 5 0. We want the point (0, 0, 1)
to be on this plane, and substituting this into the
equation above gives D 5 3. So the equation of the
>
plane containing r 5 (0, 0, 1) 1 t(1, 1, 0), tPR
and parallel to the first line is
x 2 y 2 3z 1 3 5 0.
Since (0, 1, 21) is on the first line, the distance
between these skew lines is the same as the distance
between this point and the plane just determined.
By the distance formula, this distance is
0 (0) 2 (1) 2 3(21) 1 3 0
d5
"12 1 (21)2 1 (23)2
5
5
"11
8 1.51.
8. a. –b. We will do both of these parts at once.
The two given lines are
>
r 5 (1, 22, 5) 1 s(0, 1, 21), sPR,
>
r 5 (1, 21, 22) 1 t(1, 0, 21), tPR.
By converting to parametric form, a general point
on the first line is
U(1, s 2 2, 5 2 s),
and on the second line is
V(1 1 t, 21, 22 2 t).
So the
vector
>
UV 5 (t, 1 2 s, s 2 t 2 7).
If the points U and V are those that produce the >
minimal distance between these two lines, then UV
will be perpendicular to both direction vectors,
(0, 1, 21) and (1, 0, 21). In the first case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (0, 1, 21)
5 8 2 2s 1 t
t 5 2s 2 8
In the second case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (1, 0, 21)
2
2
2
2
Calculus and Vectors Solutions Manual
5 2t 2 s 1 7
Substituting t 5 2s 2 8 into this second equation,
we get
2(2s 2 8) 2 s 1 7 5 0
s53
t 5 2s 2 8
t 5 22
Substituting these values for s and t into U and V,
we get
U(1, 1, 2)
V(21, 21, 0)
So U(1, 1, 2) is the point on the first line that
produces the minimal distance to the second line
at point V(21, 21, 0). This minimal distance is
given by
>
0 UV 0 5 0 (22, 22, 22) 0
5 "12
8 3.46
Review Exercise, pp. 552–555
1. 2x 2 y 5 31, x 1 8y 5 234, 3x 1 ky 5 38
(2x 2 y 5 31) 2 2(x 1 8y 5 234)
5 0x 2 17y 5 99
214
99
y52 ,x5
17
17
214
299
3a
b 1 ka
b 5 38
17
17
4
k52
99
1
2.
x 2 y 5 13
2 3x 1 2y 5 26
3
x 1 2y 5 219
(2 3 Equation 1 ) 1 equation 2 5 5x 1 0y 5 20
or x 5 4. Substituting x 5 4 into equation 1 gives
(4) 2 y 5 13 or y 5 29. However, when you
substitute this coordinates into the third equation,
the third equation is not consistent, so there is no
solution to this problem.
3. a. 1
x 2 y 1 2z 5 3
2 2x 2 2y 1 3z 5 1
3
2x 2 2y 1 z 5 11
Equation 2 2 equation 3 5 5z 5 210 or
z 5 22. Substituting z 5 22 into all of the equations
gives
4
x2y2453
5 2x 2 2y 2 6 5 1
6 2x 2 2y 2 2 5 11
There are no x and y variables that satisfy these
equations, so the answer is no solution.
9-25
x 1 y 1 z 5 300
x 1 y 2 z 5 98
3 x 2 y 1 z 5 100
Equation 2 1 equation 3 5 2x 5 198 x 5 99.
Substituting x 5 99 into all three equations gives:
4
y 1 z 5 201
5
y 2 z 5 21
6 2y 1 z 5 1
Equation 4 1 equation 5 5 2y 5 200 or
y 5 100. You then get z 5 101 after substituting
both x and y into equation 1 .
(99, 100, 101)
Check:
1 99 1 100 1 101 5 300
2 99 1 100 2 101 5 98
3 99 2 100 1 101 5 100
4. a. These four points will lie in the same plane if
and only if the line determined by the first two
points intersects the line determined by the last two
points. The direction vector determined by the first
two is
>
a 5 (7, 25, 1) 2 (1, 2, 6)
5 (6, 27, 25)
So these first two points determine the line with
vector equation
>
r 5 (1, 2, 6) 1 s(6,27,25), sPR.
The direction vector determined by the last two
points is
>
b 5 (23, 5, 6) 2 (1, 1, 4)
5 (24, 4, 2)
So these first two points determine the line with
vector equation
>
r 5 (1, 1, 4) 1 t(24, 4, 2), tPR.
Converting these two lines to parametric form, we
obtain the equations
1
1 1 6s 5 1 2 4t
2
2 2 7s 5 1 1 4t
3
6 2 5s 5 4 1 2t
Adding the first and second equations gives
3 2 s 5 2, so s 5 1. Substituting this into the third
equation, we get
1 5 4 1 2t
23 5 2t
So t 5 2 32. We need to check this s and t for
consistency. Substituting s 5 1 into the vector
equation for the first line gives
>
r 5 (1, 2, 6) 1 (1)(6, 27, 25)
5 (7, 25, 1)
as a point on this line. Substituting t 5 2 32 into the
vector equation for the second line gives
b.
1
2
9-26
3
>
r 5 (1, 1, 4) 1 a2 b (24, 4, 2)
2
5 (1, 1, 4) 1 (6, 26, 23)
5 (7, 25, 1)
as a point on this line. This means the two lines
intersect, and so the four points given lie in the
same plane.
b. Direction vectors for the plane containing the
four points in part a. are (6, 27, 25) and
(24, 4, 2). So a normal to this plane is
(6, 27, 25) 3 (24, 4, 2) 5 (6, 8, 24).
We will use the parallel normal (3, 4, 22). So the
equation of this plane is of the form
3x 1 4y 2 2z 1 D 5 0.
Substitute in the point (1, 2, 6) to find D.
3(1) 1 4(2) 2 2(6) 1 D 5 0
D51
The equation of the plane is
3x 1 4y 2 2z 1 1 5 0.
So, using the distance formula, this plane is distance
Z 3(0) 1 4(0) 2 2(0) 1 1 Z
d5
Z (3, 4, 22) Z
1
5
"29
8 0.19
from the origin.
5. Use the distance formula.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. The distance from A(21, 1, 2) to
3x 2 4y 2 12z 2 8 5 0
0 3(21) 2 4(1) 2 12(2) 2 8 0
d5
"32 1 (24)2 1 (212)2
53
b. The distance from B(3, 1, 22) to
8x 2 8y 1 4z 2 7 5 0
0 8(3) 2 8(1) 1 4(22) 2 7 0
d5
"82 1 (28)2 1 (4)2
1
or 0.08
5
12
>
6. r 5 (3, 1, 1) 1 t(2, 21, 2), tPR
3x 2 4y 2 5z 5 0
Find the parametric equations from the first
equation, then substitute those equations into the
second equation. Solve for t. Substitute that t-value
into the first equation.
>
r 5 (3, 1, 1) 1 t(2, 21, 2), tPR
x 5 3 1 2t
Chapter 9: Relationships Between Points, Lines, and Planes
y512t
z 5 1 1 2t
3(3 1 2t) 2 4(1 2 t) 2 5(1 1 2t) 5 0
t can be any value to satisfy this value, so the two
equations intersect along
>
r 5 (3, 1, 1) 1 t(2, 21, 2), tPR.
7. a. 1
3x 2 4y 1 5z 5 9
2
6x 2 9y 1 10z 5 9
3 9x 2 12y 1 15z 5 9
3 3 (3x 2 4y 1 5z 5 9) 5 9x 2 12y 1 15z 5 27
There is no solution because the first and third
equations are inconsistent.
1 2x 1 3y 1 4z 5 3
b.
2 4x 1 6y 1 8z 5 4
3
5x 1 y 2 z 5 1
2 3 (2x 1 3y 1 4z 5 3) 5 4x 1 6y 1 8z 5 6
There is no solution because the first and second
equations are inconsistent.
1
c.
4x 2 3y 1 2z 5 2
2
8x 2 6y 1 4z 5 4
3 12x 2 9y 1 6z 5 1
3 3 (4x 2 3y 1 2z 5 2) 5 12x 2 9y 1 6z 5 6
There is no solution because the first and third
equations are inconsistent.
8. a. 1 3x 1 4y 1 z 5 4
2 5x 1 2y 1 3z 5 2
3 6x 1 8y 1 2z 5 8
(Equation 1 ) 2 (2 3 equation 2 )
5 27x 2 5z 5 0
Letting z 5 t, then x 5 2 57 t and y 5 1 1 27 t.
5
2
x 5 2 t, y 5 1 1 t, z 5 t, tPR
7
7
1 4x 2 8y 1 12z 5 4
b.
2
2x 1 4y 1 6z 5 4
3
x 2 2y 2 3z 5 4
(Equation 1 ) 1 (4 3 equation 3 )
5 24z 5 212 or z 5 2 12. Letting z 5 2 12 creates:
4x 2 8y 5 10
2x 1 4y 5 7
(Equation 1 ) 1 (2 3 equation 2 ) 5 8x 5 24
or x 5 3. Substituting in x 5 3 and z 5 2 12 gives
y 5 14
1
1
x 5 3, y 5 , z 5 2
4
2
1
c.
x 2 3y 1 3z 5 7
2
2x 2 6y 1 6z 5 14
3 2x 1 3y 2 3z 5 27
4
5
Calculus and Vectors Solutions Manual
Letting z 5 s, then y 5 t gives x 2 3t 1 3s 5 7 or
x 5 23s 1 3t 1 7
x 5 3t 2 3s 1 7, y 5 t, z 5 s, s, tPR
9. a. 1 3x 2 5y 1 2z 5 4
1
6x 1 2y 2 z 5 2
1 6x 2 3y 1 8z 5 6
(Equation 2 ) 2 (2 3 equation 1 ) 5 12y 2 5z
5 26
Setting z 5 t,
1
5
12y 2 5t 5 26 or y 5 2 1 t
2
12
Substituting these two values into the first equation
gives x 5 12 1 361 t
1
1
21
5
x 5 1 t, y 5
1 t, z 5 t, tPR
2
36
2
12
1 2x 2 5y 1 3z 5 1
b.
2 4x 1 2y 1 5z 5 5
3 2x 1 7y 1 2z 5 4
(Equation 2 ) 2 (2 3 equation 1 )
5 12y 2 z 5 3
Setting z 5 t,
1
1
12y 2 t 5 3 or y 5 1 t
4
12
Substituting these two values into the first equation
gives x 5 98 2 31
24 t
1
9
31
1
x 5 2 t, y 5 1 t, z 5 t, tPR
8
24
4
12
10. a. 2x 1 y 1 z 5 6
x 2 y 2 z 5 29
3x 1 y 5 2
The first equation 1 the second equation gives
3x 5 23 or x 5 21. Substituting x 5 21 into the
third equation, 3(21) 1 y 5 2 or y 5 5.
Substituting these two values into the first equation,
2(21) 1 5 1 z 5 6 or z 5 3
These three planes meet at the point (21, 5, 3).
1 2x 2 y 1 2z 5 2
b.
2
3x 1 y 2 z 5 1
3 x 2 3y 1 5z 5 4
Equation 1 1 equation 2 5 5x 1 z 5 3
Equation 3 2 (3 3 equation 1 ) 5 25x 2 z
5 22.
These two equations are inconsistent, so the planes
do not intersect at any point. Geometrically the
planes form a triangular prism.
1
c.
2x 1 y 2 z 5 0
2 x 2 2y 1 3z 5 0
3 9x 1 2y 2 z 5 0
9-27
2 3 equation 1 1 equation 2 5 5x 1 z 5 0, so
z 5 25x.
Equation 3 2 equation 1 5 7x 1 y 5 0, so
y 5 27x.
Let x 5 t. The intersection of the planes is a line
through the origin with equation x 5 t, y 5 27t,
z 5 25t, tPR.
>
11. r 5 (2, 21, 22) 1 s(1, 1, 22), sPR
By substituting in different s-values, you can find
when the plane intersects the xz-plane when y 5 0
and the xy-plane when z 5 0.
The plane intersects the xz-plane at (3, 0, 24) and
the xy-plane at (1, 22, 0). Then find the distance
between these two points using the distance
formula. The distance between these two points
is 4.90.
12. a. x 2 2y 1 z 1 4 5 0
>
r 5 (3, 1, 25) 1 s(2, 1, 0), sPR
> >
m ? n 5 (2, 1, 0) ? (1, 22, 1) 5 0 Since the line’s
direction vector is perpendicular to the normal of
the plane and the point (3, 1, 25) lies on both the
line and the plane, the line is in the plane.
>
b. r 5 (7, 5, 21) 1 t(4, 3, 2), tPR
>
r 5 (3, 1, 25) 1 s(2, 1, 0), sPR
Solve for the parametric equations of both equations
and then set them equal to each other.
L1: x 5 7 1 4t, y 5 5 1 3t, z 5 21 1 2t
L2: x 5 3 1 2s, y 5 1 1 s, z 5 25
z 5 25 5 21 1 2t, t 5 22
t 5 22, x 5 21, y 5 21, z 5 25
t 5 22 corresponds to the point (21, 21, 25)
c.
x 2 2y 1 z 1 4 5 0
21 2 2(21) 1 (25) 1 4 5 0
The point (21, 21, 25) is on the plane since it
satisfies the equation of the plane.
>
d. r 5 (7, 5, 21) 1 t(4, 3, 2), tPR
(A, B, C) ? (4, 3, 2) 5 0
A 5 7, B 5 22, C 5 211
7x 2 2y 2 11z 1 D 5 0
D 5 250
7x 2 2y 2 11z 2 50 5 0
>
13. a. r 5 (3, 0, 21) 1 t(1, 1, 2), tPR
A(22, 1, 1)
x 5 3 1 t, y 5 t, z 5 21 1 2t
0 5 3 1 t 2 x, 0 5 t 2 y, 0 5 21 1 2t 2 z
"(3 1 t 2 x)2 1 (t 2 y)2 1 (21 1 2t 2 z)2
"(3 1 t 1 2)2 1 (t 2 1)2 1 (21 1 2t 2 1)2
"6t 2 1 30
t 5 0 gives the lowest distance of 5.48
9-28
b. t 5 0 corresponds to the point (3, 0, 21)
>
14. a. r 5 (1, 21, 1) 1 t(3, 2, 1), tPR
>
r 5 (22, 23, 0) 1 s(1, 2, 3), sPR
Set the equations parametric equations equal to
each other, and determine either the s or t-value.
Find the point that corresponds to this value.
L1: x 5 1 1 3t, y 5 21 1 2t, z 5 1 1 t
L2: x 5 22 1 s, y 5 23 1 2s, z 5 3s
x 5 1 1 3t 5 22 1 s
y 5 21 1 2t 5 23 1 2s
z 5 1 1 t 5 3s
s 5 0, t 5 21
s 5 0 corresponds to the point (22, 23, 0).
>
b. r 5 (1, 21, 1) 1 t(3, 2, 1), tPR
>
r 5 (22, 23, 0) 1 s(1, 2, 3), sPR
P(22, 23, 0)
S
S
n1 3 n2 5 (3, 2, 1) 3 (1, 2, 3)
5 (4, 28, 4) 5 (1, 22, 1)
>
r 5 (22, 23, 0) 1 t(1, 22, 1), tPR
15. a. Since the plane we want contains L, we can
use the direction vector for L, (1, 2, 21), as one
of the plane’s direction vectors. Since the plane
contains the point (1, 2, 23) (which is on L) and the
point K(3, 22, 4), it will contain the direction vector
(3, 22, 4) 2 (1, 2, 23) 5 (2, 24, 7)
To find a normal vector for the plane we want, take
the cross product of these two direction vectors.
(2, 24, 7) 3 (1, 2, 21) 5 (210, 9, 8)
So the plane we seek will be of the form
210x 1 9y 1 8z 1 D 5 0.
To determine the value of D, substitute in the point
(1, 2, 23) that is to be on this plane.
210(1) 1 9(2) 1 8(23) 1 D 5 0
D 5 16
The equation of the plane we seek is
210x 1 9y 1 8z 1 16 5 0.
b. Using the distance formula, the distance from
S(1, 1, 21) to the plane 210x 1 9y 1 8z 1
16 5 0 is
Z210(1) 1 9(1) 1 8(21) 1 16Z
d5
Z (210, 9, 8)Z
7
5
"245
8 0.45
16. a. 1
x1y2z51
2 2x 2 5y 1 z 5 21
3 7x 2 7y 2 z 5 k
Equation 1 1 equation 2 5 equation 4
5 3x 2 4y 5 0
Chapter 9: Relationships Between Points, Lines, and Planes
Equation 2 1 equation 3 5 equation 5
5 9x 2 12y 5 21 1 k
For the solution to this system to be a line,
equation 4 and equation 5 must be the proportional.
k 5 1 makes these two line proportional and the
solution to this system a line.
b. In part a., we found that k 5 1 by arriving at the
equivalent system
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
3
9x 2 12y 5 0
As the first and third equations are proportional,
this is really the same system as
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
Letting x 5 t in the first equation, we see that
y 5 34 t. Substituting these values for x and y into the
second equation, we find that
3
z 5 5a tb 2 2t 2 1
4
7
5 t 2 1.
4
So the direction vector for the line that solves this
system is (1, 34, 74), which is parallel to (4, 3, 7).
So equivalent parametric equations of this line are
x 5 4t
y 5 3t
z 5 21 1 7t, tPR.
So one possible vector equation of this line is
>
r 5 (0, 0, 21) 1 t(4, 3, 7), tPR.
b. In part a., we found that k 5 1 by arriving at the
equivalent system
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
3
9x 2 12y 5 0
As the first and third equations are proportional,
this is really the same system as
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
Letting x 5 t in the first equation, we see that
y 5 34t. Substituting these values for x and y into the
second equation, we find that
3
z 5 5a tb 2 2t 2 1
4
7
5 t 2 1.
4
So the direction vector for the line that solves this
system is Q 1, 34, 74 R , which is parallel to (4, 3, 7).
So equivalent parametric equations of this line are
Calculus and Vectors Solutions Manual
x 5 4t
y 5 3t
z 5 21 1 7t, tPR.
So one possible vector equation of this line is
>
r 5 (0, 0, 21) 1 t(4, 3, 7), tPR.
17. a. 1
x 1 2y 1 z 5 1
2
2x 2 3y 2 z 5 6
3 3x 1 5y 1 4z 5 5
4
4x 1 y 1 z 5 8
Equation 1 1 equation 2 5 equation 5
5 3x 2 y 5 7
(4 3 equation 2 ) 1 equation 3 5 equation 6
5 11x 2 7y 5 29
(7 3 equation 5 ) 1 equation 6
5 equation 7 5 210x 5 220y or x 5 2
Substituting into equation 5 : 6 2 y 5 7y 5 21.
Substituting into equation 1 : 2 1 22 1 z 5 1
or z 5 1.
(2, 21, 1)
1
b.
x 2 2y 1 z 5 1
2
2x 2 5y 1 z 5 21
3
3x 2 7y 1 2z 5 0
4 6x 2 14y 1 4z 5 0
Equation 2 2 (2 3 equation 1 )
5 equation 5 5 2y 2 z 5 23,
Setting z 5 t,
2y 2 t 5 23 or y 5 3 2 t
Substituting y 5 3 2 t and z 5 t into equation 1 :
x 2 2(3 2 t) 1 t 5 1 or x 5 7 2 3t
x 5 7 2 3t, y 5 3 2 t, z 5 t, tPR
9a
3c
18. 1
2 8b 1
54
b
b
3a
4c
1 4b 1
53
b
b
3a
4c
3
1 4b 2
53
b
b
a
c
x 5 , y 5 b, z 5
b
b
1
9x 2 8y 1 3z 5 4
2 23x 1 4y 1 4z 5 3
3
3x 1 4y 2 4z 5 3
3 1 2 5 8y 5 6
3
y5
4
1 9x 1 3z 5 10
2 23x 1 4z 5 0
3 3x 2 4z 5 0
1 1 3 2 5 15z 5 10
2
2
9-29
8
2
z5 ,x5
3
9
3
y 5 5 b,
4
2
8
a
a
x5 5 5 ,a5
9
b
3
3
4
1
2
c
c
z5 5 5 ,c5
3
b
3
2
4
2 3 1
a , , b
3 4 2
19. First put the equation into parametric form.
Then substitute the x, y, and z-values into
x 1 2y 2 z 1 10 5 0 to determine t. Then
substitute t back into the parametric equations
to determine the coordinates.
x11
y22
z21
5
5
5t
24
3
22
x 5 24t 2 1, y 5 3t 1 2, z 5 22t 1 1
x 1 2y 2 3z 1 10 5 0
(24t 2 1) 1 2(3t 1 2) 2 3(22t 1 1) 1 10 5 0
5
t52
4
5
5
x 5 24a2 b 2 1, y 5 3a2 b 1 2,
4
4
5
z 5 22a2 b 1 1
4
7 7
a4, 2 , b
4 2
20. Let Ar(a, b, c) denote the image point under this
reflection. We want to find a, b, and c. The equation
of the plane is x 2 y 1 z 2 1 5 0, so letting y 5 s
and z 5 t, we get x 5 1 2 t 1 s, s, tPR. These are
the parametric equations of this plane, so a general
point on this plane has coordinates P(1 2 t 1 s, s, t).
>
So PA 5 (1, 0, 4) 2 (1 2 t 1 s, s, t)
5 (t 2 s, 2s, 4 2 t)
The normal vector to this plane is (1, 21, 1), and in
>
order for PA to be perpendicular to the plane, it >
must be parallel to this normal. This means that PA
and (1, 21, 1) will have a cross product equal to
the zero vector.
(t 2 s, 2s, 4 2 t) 3 (1, 21, 1)
5 (4 2 s 2 t, 4 1 s 2 2t, 2s 2 t)
5 (0, 0, 0)
9-30
So we get the system of equations
1
42s2t50
2 4 1 s 2 2t 5 0
3
2s 2 t 5 0
Adding the first two equations gives
8 2 3t 5 0
8
t5
3
Substituting this value for t into the third equation
gives
0 5 2s 2 t
8
5 2s 2
3
4
s5
3
Substituting these values for s and t into the equation
>
for PA, we get
>
PA 5 (t 2 s, 2s, 4 2 t)
8
4 4
8
5 a 2 ,2 ,42 b
3
3 3
3
4 4 4
5 a ,2 , b
3 3 3
This is the vector that is normal to the plane, with
its head at point A(1, 0, 4) and tail at the point in
the plane
8
4 4 8
P(1 2 t 1 s, s, t) 5 Pa1 2 1 , , b
3
3 3 3
1 4 8
5 a2 , , b
3 3 3
So the
vector
>
>
PAr 5 2PA
4 4 4
5 a2 , , 2 b
3 3 3
1 4 8
5 (a, b, c) 2 a2 , , b
3 3 3
1
4
8
5 aa 1 , b 2 , c 2 b
3
3
3
This means that a 5 2 53, b 5 2 83, and c 5 2 43,
That is, the reflected point is Ar( 2 53, 83, 43) .
21. a. The first plane has normal (3, 1, 7) and the
second has normal (4, 212, 4). Their line of
intersection will be perpendicular to both of these
normals. So we can take as direction vector the
cross product of these two normals.
(3, 1, 7) 3 (4, 212, 4) 5 (88, 16, 240)
5 8(11, 2, 25)
Chapter 9: Relationships Between Points, Lines, and Planes
So let’s use (11, 2, 25) as the direction vector for
this line of intersection. To find a point on both of
these planes, solve for z in the second plane, and
substitute this into the equation for the first plane.
4x 2 12y 1 4z 2 24 5 0
4z 5 24 2 4x 1 12y
z 5 6 2 x 1 3y
0 5 3x 1 y 1 7z 1 3
5 3x 1 y 1 7(6 2 x 1 3y)
13
5 24x 1 22y 1 45
If y 5 0 in this last equation, then x 5 454 and
z 5 6 2 x 1 3y
45
562
1 3(0)
4
21
52
4
The point ( 454, 0, 214) , lies on both planes. So the
vector equation of the line of intersection for the
first two planes is
21
45
>
r 5 a , 0, 2 b 1 t(11, 2, 25), tPR.
4
4
The corresponding parametric form is
45
x5
1 11t
4
y 5 2t
21
z 5 2 2 5t, tPR.
4
We will use a similar procedure for the other two
lines of intersection. For the third plane, the normal
vector is (1, 2, 3). So a direction vector for the line
of intersection between the first and third planes is
(3, 1, 7) 3 (1, 2, 3) 5 (211, 22, 5)
5 2 (11, 2, 25)
We may use (11, 2, 25) as the direction vector for
this line of intersection. We find a point on both of
these planes in the same way as before.
x 1 2y 1 3z 2 4 5 0
x 5 4 2 2y 2 3z
0 5 3x 1 y 1 7z 1 3
5 3(4 2 2y 2 3z) 1 y 1 7z 1 3
5 26y 2 2z 1 15
Taking y 5 0 in this last equation, we get z 5 152
and
x 5 4 2 2y 2 3z
15
5 4 2 2(0) 2 3a b
2
37
52
2
Calculus and Vectors Solutions Manual
A point on both the first and third planes is
(2 372, 0, 152 ). So the vector equation for this line of
intersection is
15
37
>
r 5 a2 , 0, b 1 t(11, 2, 25), tPR,
2
2
and the corresponding parametric equations are
37
x 5 2 1 11t
2
y 5 2t
15
z5
2 5t, tPR.
2
Finally, we consider the line of intersection between
the second and third planes. In this case, a direction
vector is
(4, 212, 4) 3 (1, 2, 3) 5 (244, 28, 20)
5 24(11, 2, 25)
We may use (11, 2, 25) as the direction vector for
this line of intersection. We find a point on both of
these planes in the same way as before.
x 1 2y 1 3z 2 4 5 0
x 5 4 2 2y 2 3z
0 5 4x 2 12y 1 4z 2 24
5 4(4 2 2y 2 3z) 2 12y 1 4z 2 24
5 220y 2 8z 2 8
Taking y 5 0 in this last equation, we get z 5 21
and
x 5 4 2 2y 2 3z
5 4 2 2(0) 2 3(21)
57
A point on both the second and third planes is
(7, 0, 21). So the vector equation for this line of
intersection is
>
r 5 (7, 0, 21) 1 t(11, 2, 25), tPR,
and the corresponding parametric equations are
x 5 7 1 11t
y 5 2t
z 5 21 2 5t, PR.
b. All three lines of intersection found in part a.
have direction vector (11, 2, 25), and so they are
all parallel. Since no pair of normal vectors for
these three planes is parallel, no pair of these planes
is coincident.
2
5
3
22. 1 2 1 2 1 2 5 40
a
b
c
3
6
1
2
2 2 2 2 5 23
a2
b
c
5
4
9
3
2 2 1 2 5 67
a2
b
c
9-31
11 213
1 2 5 31
a2
b
21 229
3 14 2 5 5 5
1 2 5 55
a2
b
46
21 4 2 11 5 5 2 5 46, b 5 11, b 5 21
b
1
21
229
1
5 55, a 5 , a 5 2
2 1
a
1
2
2
1
2
5
3
1
1 1 2 5 40, c 5 , c 5 2
0.25
1
c
3
3
1
1
1
1
a 5 , a 5 2 , b 5 1, b 5 21, c 5 , c 5 2
2
2
3
3
2
2
Because each equation has each of a , b , and c2, the
possible solutions are all combinations of the
positive and negative values for a, b, and c: ( 12, 1, 13) ,
1
13
2
5
4
5
( 12, 1, 2 13) , ( 12, 21, 13) , ( 12, 21, 2 13) , ( 2 12, 1, 13) ,
( 2 12, 1, 2 13) , ( 2 12, 21, 13) , and ( 12, 21, 2 13) .
23. The general form of such a parabola is
y 5 ax 2 1 bx 1 c. We need to determine a, b, and c.
Since (21, 2), (1, 21), and (2, 1) all lie on the
parabola, we get the system of equations
1
a2b1c52
2
a 1 b 1 c 5 21
3 4a 1 2b 1 c 5 1
Adding the first and second equations gives
1
a1c5
2
Subtracting the first from the second equation gives
2b 5 23
3
b52
2
Using the fact that a 1 c 5 12 and b 5 2 32 in the
third equation gives
1 5 4a 1 2b 1 c
5 3a 1 2b 1 (a 1 c)
3
1
5 3a 1 2a2 b 1
2
2
5
5 3a 2
2
7
5 3a
2
7
a5
6
So using once more that a 1 c 5 12, we substitute
this value in for a and get
1
5a1c
2
7
1c
6
2
c52
3
So the equation of the parabola we seek is
7
3
2
y 5 x2 2 x 2 .
6
2
3
24. The equation of the plane is
4x 2 5y 1 z 2 9 5 0, which has normal (4, 25, 1).
Converting this plane to parametric form gives
x5s
y5t
z 5 9 2 4s 1 5t, s, tPR.
So for any point Y(s, t, 9 2 4s 1 5t) on this plane,
we >can form the vector
XY 5 (s, t, 9 2 4s 1 5t) 2 (3, 2, 25)
5 (s 2 3, t 2 2, 14 2 4s 1 5t)
This vector is perpendicular to the plane when it
is parallel to the normal vector (4, 25, 1). Two
vectors are parallel precisely when their cross
product is the zero vector.
(s 2 3, t 2 2, 14 2 4s 1 5t) 3 (4, 25, 1)
5 (68 1 26t 2 20s, 59 1 20t 2 17s, 23 2 4t 2 5s)
5 (0, 0, 0)
So we get the system of equations
1 68 1 26t 2 20s 5 0
2 59 1 20t 2 17s 5 0
3
23 2 4t 2 5s 5 0
Subtracting four times the third equation from the
first equation gives
42t 2 24 5 0
4
t5
7
Substituting this value for t into the second equation
gives
0 5 59 1 20t 2 17s
4
5 59 1 20a b 2 17s
7
493
17s 5
7
29
s5
7
Substituting these values for s and t into the equation
for Y gives
29
29 4
Y(s, t, 9 2 4s 1 5t) 5 Ya , , 9 2 4a b
7 7
7
4
29 4 33
1 5a bb 5 a , , 2 b
7
7 7
7
5
So the point M we wanted is M( 297, 47, 2 337 ).
9-32
Chapter 9: Relationships Between Points, Lines, and Planes
11x 2 2 14x 1 9
A
Bx 1 C
5
1 2
2
(3x 2 1)(x 1 1)
3x 2 1
x 11
11x 2 2 14x 1 9
(3x 2 1)(x 2 1 1)
A(x 2 1 1) 1 (Bx 1 C) 3x 2 1
5
(3x 2 1)(x 2 1 1)
11x 2 2 14x 1 9 5 (A 1 3B)x 2 1 (3C 2 B)x
1 (A 2 C)
A 2 C 5 9, 3C 2 B 5 214, A 1 3B 5 11
B 5 3C 1 14, A 5 C 1 9
A 1 3(3C 1 14) 5 11, A 1 9C 5 231
(C 1 9) 1 9C 5 231
10C 5 240, C 5 24
B 5 3(24) 1 14 5 2, A 5 (24) 1 9 5 5
A 5 5, B 5 2, C 5 24
26.> a. The vector
EF 5 (21, 24, 26) 2 (4, 0, 3)
5 (25, 24, 23)
This is a direction vector for the line containing the
segment EF. The point E(21, 24, 26) is on this
line, so the vector equation of this line is
>
r 5 (21, 24, 26) 1 t(25, 24, 23), tPR.
b. Based on the equation of the line found in part a.,
a general point on this line is of the form
J(21 2 5t, 24 2 4t, 26 2 3t), tPR.
For> this general point, the vector
JD 5 (3, 0, 7) 2 (21 2 5t, 24 24t, 26 2 3t)
5 (4 1 5t, 4 1 4t, 13 1 3t)
This vector will be perpendicular to the direction
vector for the line found in part a. at the point J we
seek. This means that
0 5 (4 1 5t, 4 1 4t, 13 1 3t) ? (25, 24, 23)
5 25(4 1 5t) 2 4(4 1 4t) 2 3(13 1 3t)
5 275 2 50t
3
t52
2
Substituting this value of t into the equation for the
general point on the line in part a.,
J(21 2 5t, 24 2 4t, 26 2 3t)
3
3
3
5 Ja21 2 5a2 b, 24 2 4a2 b, 26 2 3a2 bb
2
2
2
13
3
5 a , 2, 2 b
2
2
These are the coordinates for the point J we wanted.
c. Using the coordinates for J found in part b.,
>
13
3
JD 5 (3, 0, 7) 2 a , 2, 2 b
2
2
7
17
5 a2 , 22, b
2
2
25.
Calculus and Vectors Solutions Manual
This vector forms the height of ^ DEF, and the
length of this vector is
>
7
17
@ JD @ 5 ` a2 , 22, b `
2
2
2
7
17 2
5 a2 b 1 (22)2 1 a b
Å
2
2
177
Å 72
8 9.41
The length of the base of ^ DEF is
>
@ EF @ 5 0 (25, 24, 23) 0
5
5 "(25)2 1 (24)2 1 (23)2
5 "50
8 7.07
So the area of ^ DEF equals
177
5
1
( !50) a
b 5 !177
2
Å 2
2
8 33.26 units2
27. 3x 2 2z 1 1 5 0
4x 1 3y 1 7 5 0
(5, 25, 5)
>
>
n1 3 n2 5 (3, 0, 22) 3 (4, 3, 0) 5 (6, 28, 9)
6x 2 8y 1 9z 1 D 5 0
D 5 2115
6x 2 8y 1 9z 2 115 5 0
Chapter 9 Test, p. 556
>
1. a. r 1 5 (4, 2, 6) 1 s(1, 3, 11), sPR,
>
r 2 5 (5, 21, 4) 1 t(2, 0, 9), tPR
L1: x 5 4 1 s, y 5 2 1 3s, z 5 6 1 11s
L2: x 5 5 1 2t, y 5 21, z 5 4 1 9t
y 5 21 5 2 1 3s
s 5 21
L1: x 5 4 1 (21), y 5 2 1 3(21),
z 5 6 1 11(21)
x 5 3, y 5 21, z 5 25
(3, 21, 25)
b.
x2y1z1150
3 2 (21) 1 (25) 1 1 5 0
311251150
050
2. Use the distance equation.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. A(3, 2, 3)
8x 2 8y 1 4z 2 7 5 0
9-33
d5
5
0 8x0 2 8y0 1 4z0 2 7 0
"(8) 1 (28) 1 (4)
0 8(3) 2 8(2) 1 4(3) 2 7 0
2
2
Equation 2 1 (2 3 equation
4x 1 y 5 21
Equation 2 1 (8 3 equation
2
"(8)2 1 (28)2 1 (4)2
13
5
or 1.08
12
b. First, find any point on one of the planes, then use
the other plane equation with the distance formula.
2x 2 y 1 2z 2 16 5 0
2x 2 y 1 2z 1 24 5 0
2(8) 2 (0) 1 2(0) 2 16 5 0
A(8, 0, 0)
0 2x0 2 1y0 1 2z0 1 24 0
d5
"(2)2 1 (21)2 1 (2)2
0 2(8) 2 1(0) 1 2(0) 1 24 0
5
"(2)2 1 (21)2 1 (2)2
40
5
or 13.33
3
3. a. L1: 2x 1 3y 2 z 5 3
L2: 2x 1 y 1 z 5 1
L1 1 2L2: 5y 1 z 5 5
z 5 t,
5y 1 (t) 5 5
t
y512
5
2x 1 y 1 z 5 1
t
2x 1 a1 2 b 1 (t) 5 1
5
4t
x5
5
t
4t
x 5 , y 5 1 2 , z 5 t, tPR
5
5
b. To determine the point of intersection with the
xz-plane, set the above y parametric equation equal
to 0 and solve for the t. This t corresponds to the
point of intersection.
t
4t
x 5 , y 5 1 2 , z 5 t, tPR
5
5
t
0512
5
t55
(5)
4(5)
x5
,y512
, z 5 (5), tPR
5
5
(4, 0, 5)
4. a. 1
x 2 y 1 z 5 10
2 2x 1 3y 2 2z 5 221
2
1
1
1
3
x1 y1 z52
2
5
4
2
9-34
6x 1
1)
5
3)
5
31
y 5 225
5
31
(4x 1 y 5 21)
5
31
1 a6x 1 y 5 225b
5
218.8x 5 218.8
x51
4(1) 1 y 5 21
y 5 25
(1) 2 (25) 1 z 5 10
z54
(1, 25, 4)
b. The three planes intersect at this point.
1
5. a.
x 2 y 1 z 5 21
2 2x 1 2y 2 z 5 0
3 x 2 5y 1 4z 5 23
Equation 2 1 (2 3 equation 1 ) 5
4x 1 z 5 22
4x 1 z 5 22
z5t
4x 1 (t) 5 22
1
t
x52 2
2
4
x 2 y 1 z 5 21
1
t
a2 2 b 2 y 1 (t) 5 21
2
4
3t
1
y5 1
4
2
3t
1
t
1
x 5 2 2 , y 5 1 , z 5 t, tPR
2
4
4
2
b. The three planes intersect at this line.
6. a. L1: x 1 y 1 z 5 0
L2: x 1 2y 1 2z 5 1
L3: 2x 2 y 1 mz 5 n
L2 1 2L3: 5x 1 0y 1 (2m 1 2)z 5 2n 1 1
L1 1 L3: 3x 1 0y 1 (m 1 1)z 5 n
5
(3x 1 0y 1 (m 1 1)z 5 n)
3
5
5
5 5x 1 0y 1 (m 1 1)z 5 n
3
3
Then set the two new equations to each other and
solve for a m and n value that would give equivalent
equations.
2
Chapter 9: Relationships Between Points, Lines, and Planes
5
n
3
2n 1 1
5
2m 1 2 5 (m 1 1)
3
m 5 21
5
n 5 2n 1 1
3
n 5 23
b. L1: x 1 y 1 z 5 0
L2: x 1 2y 1 2z 5 1
L3: 2x 2 y 2 z 5 23
L1 1 L2: 3x 5 23, x 5 21
(21) 1 y 1 z 5 0
z5t
(21) 1 y 1 (t) 5 0
y512t
x 5 21, y 5 1 2 t, z 5 t, tPR
7. First find the parametric equations of each line.
Then set these equations equal to each other to find
a set of new equations. Use the dot product to
determine another set of equations that you will
solve for t and s. Find the corresponding points to
these values and the distance between them, which
is the distance between the two lines.
>
L1: r 5 (21, 23, 0) 1 s(1, 1, 1), sPR
>
L2: r 5 (25, 5, 28) 1 t(1, 2, 5), tPR
L1: x 5 21 1 s, y 5 23 1 s, z 5 s
L2: x 5 25 1 t, y 5 5 1 2t, z 5 28 1 5t
>
UV 5 3 (21 1 s) 2 (25 1 t), (23 1 s)
2 (5 1 2t), s 2 (28 1 5t)4
>
UV 5 (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t)
>
m1 ? UV 5 0
(1, 1, 1) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
(1, 2, 5) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
L4: 4 1 3s 2 8t 5 0,
L5: 28 1 8s 2 30t 5 0
8 3 L1 1 (23) 3 L2 yields
32 1 24s 2 64t 2 84 2 24s 1 90t 5 0, so t 5 2.
Then s 5 4. The points corresponding to these
values of s and t are (21, 3, 0) 1 4(1, 1, 1)
5 (3, 1, 4) and (25, 5, 28) 1 2(1, 2, 5)
5 (23, 9, 2).
5
5x 1 0y 1 (m 1 1)z 5
3
5x 1 0y 1 (2m 1 2)z 5
d 5 "(3 2 (23))2 1 (1 2 9)2 1 (4 2 2)2
5 "(6)2 1 (28)2 1 (2)2
5 "36 1 64 1 4
5 "104 or 10.20
Calculus and Vectors Solutions Manual
Cumulative Review of Vectors,
pp. 557–560
1. a. The angle, u, between the two
vectors is found
>
>
a?b
from the equation cos (u) 5 > > .
@a @ @b@
> >
a ? b 5 (2, 21, 22) ? (3, 24, 12)
5 2(3) 2 1(24) 2 2(12)
5 214
>
0 a 0 5 "22 1 (21)2 1 (22)2
53
0 b 0 5 "32 1 (24)2 1 122
5 13
So u 5 cos21 ( 3 214
3 13 )
8 111.0°
>
>
>
b. The scalar projection of a on b is equal to
>
0 a 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 13
>
>
>
and 0 a 0 5 3, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 3 5 2 13 . The vector projection of a on b
is equal to the scalar projection multiplied by the unit
>
vector in the direction of b. So the vector projection
1
52 56
, , 2 168
is 2 14
13 3 13 (3, 24, 12) 5 (2
169 ).
> 169 169
>
c. > The scalar projection of b on a is equal to
0 b 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 > 13
>
>
and 0 b 0 5 13, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 13 5 2 3 . The vector projection of b on a
is equal to the scalar projection multiplied by the
>
unit vector in the direction of a . So the vector
projection is 2 143 3 13 (2, 21, 22) 5 (2 289, 149, 289 ).
2. a. Since the normal of the first plane is (4, 2, 6)
and the normal of the second is (1, 21, 1), which
are not scalar multiples of each other, there is a line
of intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The first equation minus four times the second
equation yields 0x 1 6y 1 2z 1 6 5 0. We may
divide by two to simplify, so 3y 1 z 1 3 5 0. If we
let y 5 t, then 3t 1 z 1 3 5 0, or z 5 23 2 3t.
Substituting these into the second equation yields
x 2 (t) 1 (23 2 3t) 2 5 5 0 or x 5 8 1 4t.
So the equation of the line in parametric form is
x 5 8 1 4t, y 5 t, z 5 23 2 3t, tPR.
9-35
To check that this is correct, we substitute in the
solution to both initial equations
4x 1 2y 1 6z 2 14 5 4(8 1 4t) 1 2(t)
1 6(23 2 3t) 2 14
50
and x 2 y 1 z 2 5
5 (8 1 4t) 2 (t) 1 (23 2 3t) 2 5
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The angle between two planes is the same as the
angle between their corresponding normal vectors.
0 (4, 2, 6)0 5 "42 1 22 1 62
5 !56
0 (1, 21, 1) 0 5 "12 1 12 1 12
5 !3
(4, 2, 6) ? (1, 21, 1) 5 8, so the angle between the
planes is cos21 Q !38!56 R 8 51.9°.
> >
>
x?y
3. a. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since x
>
>
>
and y are unit vectors, 0 x 0 5 0 y 0 5 1, and moreover
> >
> >
1
x?y
1
cos (60°) 5 . So x ? y 5
5 .
2
131
2
b. Scalar multiples can be brought out to the front
>
>
> >
of dot products. Hence 2x ? 3y 5 (2)(3)(x ? y ),
>
>
and so by part a., 2x ? 3y 5 2 3 3 3 12 5 3.
c. The dot product is distributive,
>
>
>
>
so (2x 2 y ) ? (x 1 3y )
>
>
>
>
>
>
5 2x ? (x 1 3y ) 2 y ? (x 1 3y )
> >
>
>
> >
>
>
5 2x ? x 1 2x ? 3y 2 y ? x 2 y ? 3y
> >
>
>
> >
> >
5 2x ? x 1 2x ? 3y 2 x ? y 2 3y ? y
>
>
> >
> >
Since x and y are unit vectors, x ? x 5 y ? y 5 1,
and so by using the values found in part a. and b.,
>
>
>
>
(2x 2 y ) ? (x 1 3y ) 5 2(1) 1 (3) 2 A 12 B 2 3(1)
3
5
2 >
>
>
>
>
>
>
>
4. a. 2(i 2 2j 1 3k ) 2 4(2i 1 4j 1 5k ) 2 (i 2 j )
>
>
>
>
>
>
>
>
5 2i 2 4j 1 6k 2 8i 2 16j 2 20k 2 i 1 j
>
>
>
5 27i 2 19j 2 14k
>
>
>
>
>
>
>
>
b. 22(3i> 2 4j> 2 5k>) ? (2i> 1 3k> ) 1 2i
?
(3j
2
2k
)
>
5 22(3i 2 4j 2 5k ) ? (2i 1 0j 1 3k )
>
>
>
>
>
>
1 2(i 1 0j 1 0k ) ? (0i 1 3j 2 2k )
5 22(3(2) 2 4(0) 2 5(3)) 1 2(1(0)
1 0(3) 1 0(22))
5 22(29) 1 2(0)
5 18
9-36
5. The direction vectors for the positive x-axis,
y-axis, and z-axis are (1, 0, 0), (0, 1, 0), and (0, 0, 1),
respectively.
0 (4, 22, 23) 0 5 "42 1 (22)2 1 (23)2
5 !29,
and 0 (1, 0, 0) 0 5 0 (0, 1, 0) 0
5 0 (0, 0, 1) 0
5 !1
5 1.
(4, 22, 23) ? (1, 0, 0) 5 4, so the angle the vector
4
makes with the x-axis is cos21 Q 1 !29
R 8 42.0°.
(4, 22, 23) ? (0, 1, 0) 5 22, so the angle the vector
makes with the y-axis is cos21 Q 1 22
!29 R 5 111.8°.
(4,22,23) ? (0, 0, 1) 5 23, hence the angle the
vector makes with the z-axis is cos21 Q 1 23
!29 R 8 123.9°.
>
>
6. a. a 3 b 5 (1, 22, 3) 3 (21, 1, 2)
5 (22(2) 2 3(1), 3(21) 2 1(2),
1(1) 2 (22)(21))
5 (27, 25, 21)
b. By the> scalar law for> vector multiplication,
>
>
2a 3 3b 5 2(3)(a 3
> b)
>
5 6(a 3 b )
5 6(27, 25, 21) 5 (242, 230, 26)
>
c.> The area of a parallelogram determined by a and
b is equal
> to the magnitude of the cross product of
>
a and b.
A 5 area of> parallelogram
>
5 0a 3 b0
5 0 (27, 25, 21) 0
5 "(27)2 1 (25)2 1 (21)2
8 >8.66 square units >
>
>
d. (b 3 a ) 5 2 (a 3 b )
5 2 (27, 25, 21)
5 (7, 5, 1)
>
>
>
So c ? (b 3 a ) 5 (3, 24, 21) ? (7, 5, 1)
5 3(7) 2 4(5) 2 1(1)
50
>
>
7. A unit vector perpendicular to both a and b can
be determined
> from
> any vector perpendicular to
>
>
both a and b>. a 3 b is a vector perpendicular to
>
both a and
b.
>
>
a 3 b 5 (1, 21, 1) 3 (2, 22, 3)
5 (21(3) 2 1(22), 1(2) 2 1(3),
1(22) 2 (21)(2))
5 (21, 21, 0)
Chapter 9: Relationships Between Points, Lines, and Planes
>
>
0 a 3 b 0 5 0 (21, 21, 0) 0
5 "(21)2 1 (21)2 1 02
5 !2
1
1
1
So !2
(21, 21, 0) 5 Q 2 !2
, 2 !2
, 0 R is an unit vector
>
>
1
1
, !2
, 0 R is another.
perpendicular to both a and b. Q !2
8. a. Answers may vary. For example:
>
A direction
vector for the line is AB.
>
AB 5 (1, 2, 3) 2 (2, 23, 1)
5 (21, 5, 2)
Since A(2, 23, 1) is a point on the line,
>
r 5 (2, 23, 1) 1 t(21, 5, 2), tPR, is a vector
equation for a line and the corresponding parametric
equation is x 5 2 2 t, y 5 23 1 5t, z 5 1 1 2t,
tPR.
b. If the x-coordinate of a point on the line is 4, then
2 2 t 5 4, or t 5 22. At t 5 22, the point on the
line is (2, 23, 1) 2 2(21, 5, 2) 5 (4, 213, 23).
Hence C(4, 213, 23) is a point on the line.
9. The direction vector of the first line is (21, 5, 2),
while the direction vector for the second line is
(1, 25, 22) 5 2 (21, 5, 2). So the direction vectors
for the line are collinear. Hence the lines are parallel.
The lines coincide if and only if for any point on
the first line and any point on the second line, the
vector connecting the two points is a multiple of the
direction vector for the lines.
(2, 0, 9) is a point on the first line and (3, 25, 10) is
a point on the second line.
(2, 0, 9) 2 (3, 25, 10) 5 (21, 5, 21) 2 k(21, 5, 2)
for any kPR. Hence the lines are parallel and distinct.
10. The direction vector for the parallel line is
(0, 1, 1). Since parallel lines have collinear direction
vectors, (0, 1, 1) can be used as a direction vector
for the line. Since (0, 0, 4) is a point on the line,
>
r 5 (0, 0, 4) 1 t(0, 1, 1), tPR, is a vector equation
for a line and the corresponding parametric equation
is x 5 0, y 5 t, z 5 4 1 t, tPR.
11. The line is parallel to the plane if and only if the
direction vector for the line is perpendicular to the
normal vector for the plane. The normal vector for
the plane is (2, 3, c). The direction vector for the
line is (2, 3, 1). The vectors are perpendicular if and
only if the dot product between the two is zero.
(2, 3, c) ? (2, 3, 1) 5 2(2) 1 3(3) 1 c(1)
5 13 1 c
So if c 5 213, then the dot product of normal
vector and the direction vector is zero. Hence for
c 5 213, the line and plane are parallel.
Calculus and Vectors Solutions Manual
12. First put the line in its corresponding parametric
form. (3, 1, 5) is a direction vector and (2, 25, 3) is
the origin point, so a parametric equation for the
line is x 5 2 1 3s, y 5 25 1 s, z 5 3 1 5s, sPR.
If we substitute these coordinates into the equation
of the plane, we may find the s value where the line
intersects the plane.
5x 1 y 2 2z 1 2
5 5(2 1 3s) 1 (25 1 s) 2 2(3 1 5s) 1 2
5 10 1 15s 1 2 5 1 s 2 6 2 10s 1 2
5 1 1 6s
So if 5x 1 y 2 2z 1 2 5 0, then 1 1 6s 5 0 or
s 5 2 16. At s 5 2 16, the point on the line is ( 32, 2 316, 136) .
To check that this point is also on the plane, we
substitute the x, y, z values into the plane equation
and check that it equals zero.
3
31
13
5x 1 y 2 2z 1 2 5 5a b 1 a2 b 2 2a b 1 2
2
6
6
50
Hence ( 32, 2 316, 136) is the point of intersection between
the line and the plane.
13. a.
z
(0, 0, 3)
(0, 3, 0)
x
y
(6, 0, 0)
Two direction vectors are:
(0, 3, 0) 2 (0, 0, 3) 5 (0, 3, 23)
and
(6, 0, 0) 2 (0, 0, 3) 5 (6, 0, 23).
9-37
b.
z
(–3, –2, 2)
(0, 0, 0)
y
(3, 2, 1)
x
Two direction vectors are:
(23, 22, 2) 2 (0, 0, 0) 5 (23, 22, 2)
and
(3, 2, 1) 2 (0, 0, 0) 5 (3, 2, 1).
c.
z
(0, 3, 6)
(0, 0, 0)
x
(1, 1, –1)
y
Two direction vectors are:
(0, 3, 6) 2 (0, 0, 0) 5 (0, 3, 6)
and
(1, 1, 21) 2 (0, 0, 0) 5 (1, 1, 21).
14. The plane is the right bisector joining
P(1, 22, 4) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (2, 23, 24). So the line connecting the
two points is (1, 22, 4) 1 t(2, 23, 24), tPR, or in
9-38
corresponding parametric form is x 5 1 1 2t,
y 5 22 2 3t, z 5 4 2 4t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we substitute the parametric equation into the plane
equation and solve for t.
2x 2 3y 2 4z 1 66
5 2(1 1 2t) 2 3(22 2 3t) 2 4(4 2 4t) 1 66
5 2 1 4t 1 6 1 9t 2 16 1 16t 1 66
5 58 1 29t
So if 2x 2 3y 2 4z 1 66 5 0, then 58 1 29t 5 0,
or t 5 22.
So the point of intersection is occurs at t 5 22, since
the origin point is P and the intersection occurs at the
midpoint of the line connecting P and its image, the
image point occurs at t 5 2 3 (22) 5 24.
So the image point is at x 5 1 1 2(24) 5 27,
y 5 22 2 3(24) 5 10, z 5 4 2 4(24) 5 20.
So the image point is (27, 10, 20).
15. Let (a, b, c) be the direction vector for this line.
>
So a line equation is r 5 (1, 0, 2) 1 t(a, b, c), tPR.
Since (1, 0, 2) is not on the other line, we may
choose a, b, and c such that the intersection occurs
at t 5 1. Since the line is supposed to intersect the
given line at a right angle, the direction vectors
should be perpendicular. The direction vectors are
perpendicular if and only if their dot product is zero.
The direction vector for the given line is (1, 1, 2).
(a, b, c) ? (1, 1, 2) 5 a 1 b 1 2c 5 0, so
b 5 2a 2 2c.
Also (1, 0, 2) 1 (a, b, c) 5 (1 1 a, b, 2 1 c) is the
point of intersection.
By substituting for b,
(1 1 a, b, 2 1 c) 5 (1 1 a, 2a 2 2c, 2 1 c).
So for some s value,
x 5 22 1 s 5 1 1 a
y 5 3 1 s 5 2a 2 2c
z 5 4 1 2s 5 2 1 c
Subtracting the first equation from the second yields
the equation, 5 1 0s 5 22a 2 2c 2 1.
Simplifying this gives 6 5 22a 2 2c or just
a 1 c 5 23.
Subtracting twice the first equation from the third
yields the equation, 8 5 22a 1 c.
So a 1 c 5 23 and 22a 1 c 5 8, which is two
equations with two unknowns. Twice the first plus
the second equations gives 0a 1 3c 5 2 or c 5 23.
Solving back for a gives 2 113 and since b 5 2a 2 2c,
b 5 73. Since a 1 b 1 2c 5 0, the direction vectors,
Chapter 9: Relationships Between Points, Lines, and Planes
(1, 1, 2) and (a, b, c) are perpendicular. A direction
vector for the line is (211, 7, 2).
We need to check that
7 8
(1, 0, 2) 1 (a, b, c) 5 ( 28
3 , 3 , 3 ) is a point on the
given line.
x 5 22 1 s 5 2 83, at s 5 2 23. The point on the given
7 8
line at s 5 2 23 is Q 28
3 , 3 , 3 R . Hence
>
q 5 (1, 0, 2) 1 t(211, 7, 2), tPR, is a line that
intersects the given line at a right angle.
16. a. The Cartesian equation is found by taking
> the
cross
> product of the two direction vectors, AB and
AC.
>
AB 5 (22, 0, 0) 2 (1, 2, 3)
> 5 (23, 22, 23)
AC 5 (1, 4, 0) 2 (1, 2, 3) 5 (0, 2, 23)
>
>
AB 3 AC 5 (22(23) 2 (23)(2),
23(0) 2 (23)(23),
23(2) 2 (22)(0))
5 (12, 29, 26)
So 5 (12, 29, 26) is a normal vector for the
plane, so the plane has the form
12x 2 9y 2 6z 1 D 5 0, for some constant D. To
find D, we know that A(1, 2, 3) is a point on the
plane, so 12(1) 2 9(2) 2 6(3) 1 D 5 0. So
224 1 D 5 0, or D 5 24. So the Cartesian
equation for the plane is 12x 2 9y 2 6z 1 24 5 0.
b. Substitute into the formula to determine distance
between a point and a plane. So the distance, d, of
(0, 0, 0) to the plane 12x 2 9y 2 6z 1 24 5 0 is
equal to
@ 12 (0) 2 9 (0) 2 6 (0) 1 24 @
"122 1 (29)2 1 (26)2
24
!261 8 1.49.
c. Since the plane is perpendicular to the z-axis, a
direction vector for the z-axis acts as a normal vector
for the plane. Hence (0, 0, 1) is a normal vector for
the plane. So the plane has the form z 1 D 5 0, for
some constant D. To find D, we know that (3, 21, 3)
is a point on the plane, so
0(3) 1 0(21) 1 (3) 1 D 5 0. So 3 1 D 5 0, or
D 5 23. So the Cartesian equation for the plane is
z 2 3 5 0.
d. The Cartesian equation can be found by taking
the cross product of the two direction vectors for
the plane. Since (3, 1, 22) and (1, 3, 21) are two
points on the plane
(3, 1, 22) 2 (1, 3, 21) 5 (2, 22, 21) is a
direction vector for the plane. Since the plane is
parallel to the y-axis, (0, 1, 0) is also a direction
vector for the plane.
(2, 22, 21) 3 (0, 1, 0) 5 (22(0) 2
(21)(1), (21)(0)2 (2)(0), 2(1) 2 (22)(0))
5 (1, 0, 2)
So (1, 0, 2) is a normal vector for the plane, so the
plane has the form x 1 0y 1 2z 1 D 5 0, for some
constant D. To find D, we know that (3, 1, 22) is a
point on the plane, so
(3) 1 0(1) 1 2(22) 1 D 5 0. So 21 1 D 5 0,
or D 5 1. So the Cartesian equation for the plane is
x 1 2z 1 1 5 0.
18.
E
100 km/h
45°
F
.
So d 5
17. a. (3, 25, 4) is a normal vector for the plane, so
the plane has the form 3x 2 5y 1 4z 1 D 5 0, for
some constant D. To find D, we know that
A(21, 2, 5) is a point on the plane, so
3(21) 2 5(2) 1 4(5) 1 D 5 0. So 7 1 D 5 0,
or D 5 27. So the Cartesian equation for the plane
is 3x 2 5y 1 4z 2 7 5 0.
b. Since the plane is perpendicular to the line
connecting (2, 1, 8) and (1, 2, 24), a direction
vector for the line acts as a normal vector for the
plane. So (2, 1, 8) 2 (1, 2, 24) 5 (1, 21, 12) is a
normal vector for the plane. So the plane has the
form x 2 y 1 12z 1 D 5 0, for some constant D.
To find D, we know that K(4, 1, 2) is a point on the
plane, so (4) 2 (1) 1 12(2) 1 D 5 0. So
27 1 D 5 0, or D 5 227. So the Cartesian
equation for the plane is x 2 y 1 12z 2 27 5 0.
Calculus and Vectors Solutions Manual
400 km/h
45°
100 km/h
Position Diagram
R
400 km/h
D
Vector Diagram
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 4002 1 1002 2 2(400)(100) cos (45°)
8 336.80 km> h.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin 45°
sin /EDF
8
.
336.80
100
9-39
sin 45°
3 100.
336.80
sin /EDF 8 0.2100.
Thus /EDF 8 12.1°, so the resultant velocity is
336.80 km> h, N 12.1° W.
19. a. The simplest way is to find the parametric
equation, then find the corresponding vector equation.
If we substitute x 5 s and y 5 t and solve for z, we
obtain 3s 2 2t 1 z 2 6 5 0 or z 5 6 2 3s 1 2t.
This yields the parametric equations x 5 s, y 5 t,
and z 5 6 2 3s 1 2t. So the corresponding vector
>
equation is r 5 (0, 0, 6) 1 s(1, 0, 23) 1 t(0, 1, 2),
s, tPR. To check that this is correct, find the
Cartesian equation corresponding to the above
vector equation and see if it is equivalent to the
Cartesian equation given in the problem. A normal
vector to this plane is the cross product of the two
directional
vectors.
>
n 5 (1, 0, 23) 3 (0, 1, 2) 5 (0(2) 2 (23)(1),
23(0) 2 1(2), 1(1) 2 0(0))
5 (3, 22, 1)
So (3, 22, 1) is a normal vector for the plane, so the
plane has the form 3x 2 2y 1 z 1 D 5 0, for some
constant D. To find D, we know that (0, 0, 6) is a point
on the plane, so 3(0) 2 2(0) 1 (6) 1 D 5 0.
So 6 1 D 5 0, or D 5 26. So the Cartesian equation
for the plane is 3x 2 2y 1 z 2 6 5 0. Since this is
the same as the initial Cartesian equation, the vector
equation for the plane is correct.
b.
z
sin /EDF 8
(0, 0, 6)
(0, –3, 0)
vector for the plane. The direction vector of the line
is (2, 21, 2) and the normal vector for the plane
is (1, 2, 1).
0 (2, 21, 2) 0 5 "22 1 (21)2 1 22
5 !9
5 3.
0 (1, 2, 1) 0 5 "12 1 22 1 12
5 !6
(2, 21, 2) ? (1, 2, 1) 5 2(1) 2 1(2) 1 2(1) 5 2
So the angle between the normal vector and the
2
direction vector is cos21 Q 3 !6
R 8 74.21°. So
u 8 90° 2 74.21° 5 15.79°.
To the nearest degree, u 5 16°.
b. The two planes are perpendicular if and only if
their normal vectors are also perpendicular.
A normal vector for the first plane is (2, 23, 1) and
a normal vector for the second plane is
(4, 23, 217). The two vectors are perpendicular if
and only if their dot product is zero.
(2, 23, 1) ? (4, 23, 217) 5 2(4) 2 3(23)
1 1(217)
5 0.
Hence the normal vectors are perpendicular. Thus
the planes are perpendicular.
c. The two planes are parallel if and only if their
normal vectors are also parallel. A normal vector for
the first plane is (2, 23, 2) and a normal vector for
the second plane is (2, 23, 2). Since both normal
vectors are the same, the planes are parallel. Since
2(0) 2 3(21) 1 2(0) 2 3 5 0, the point
(0, 21, 0) is on the second plane. Yet since
2(0) 2 3(21) 1 2(0) 2 1 5 2 2 0, (0, 21, 0) is
not on the first plane. Thus the two planes are
parallel but not coincident.
21.
25 N
y
x
60°
(2, 0, 0)
40 N
Position diagram
E
40 N
F
120°
20. a. The angle, u, between the plane and the line
is the complementary angle of the angle between
the direction vector of the line and the normal
9-40
25 N
60°
D
R
120°
40 N
Vector diagram
Chapter 9: Relationships Between Points, Lines, and Planes
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 402 1 252 2 2(40)(25) cos (120°)
8 56.79 N.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin 120°
sin /EDF
8
.
56.79
40
sin 120°
sin /EDF 8
3 40.
56.79
sin /EDF 8 0.610.
Thus /EDF 8 37.6°, so the resultant force is
approximately 56.79 N, 37.6° from the 25 N force
towards the 40 N force. The equilibrant force has
the same magnitude as the resultant, but it is in
the opposite direction. So the equilibrant is
approximately 56.79 N, 180° 2 37.6° 5 142.4°
from the 25 N force away from the 40 N force.
22.
a
b
–b
–b
a
a –b
b.
1
2
2a
1
2
b
2a +
1
2
b
b
2a
>
23. a. The unit vector in the same direction of a is
>
>
simply a divided by the magnitude of a .
>
0 a 0 5 "62 1 22 1 (23)2
5 "49
57
>
So the unit vector in the same direction of a is
1 >
> a 5 17 (6, 2, 23) 5 ( 67, 27, 2 37 ).
0a0
>
b. The unit vector in the opposite direction of a is
simply the negative of the unit vector found in part
a. So the vector is 2 A 67, 27, 2 37 B 5 A2 67, 2 27, 37 B.
24. a. Since OBCD is a parallelogram, the point
C
>
occurs at (21, 7) 1 (9, 2) 5 (8, 9). So> OC is one
vector
equivalent to a diagonal and BD is the other.
>
OC> 5 (8, 9) 2 (0, 0) 5 (8, 9)
BD 5 (9, 2) 2 (1, 7) 5 (10, 25)
Calculus and Vectors Solutions Manual
0 (8, 9) 0 5 "82 1 92
b.
5 "145
0 (10, 25) 0 5 "102 1 (25)2
5 "125
(8, 9) ? (10, 25) 5 8(10) 1 9(25)
5 235
So the angle between these diagonals is
235
cos21 A !145
!125 B 8 74.9°.
>
>
c. OB 5 (21, 7) and OD 5 (9, 2)
0 (21, 7) 0 5 "(21)2 1 72.
5 "50
0 (9, 2) 0 5 "92 1 22
5 "85
(21, 7) ? (9, 2) 5 2 (9) 1 7(2)
55
So the angle between these diagonals is
cos21 A !505!85 B 8 85.6°.
25. a. First step is to use the first equation to
remove x from the second and third.
1 x2y1z52
2 2x 1 y 1 2z 5 1
3 x 2 y 1 4z 5 5
So we have
4 0x 1 0y 1 3z 5 3, 1 1 2
5 0x 1 0y 1 3z 5 3, 21 3 1 1 3
Hence 3z 5 3, or z 5 1. Since both equations are
the same, this implies that there are infinitely many
solutions. Let x 5 t, then by substituting into the
equation 2, we obtain
2t 1 y 1 2(1) 5 1, or y 5 21 1 t.
Hence the solution to these equations is x 5 t,
y 5 21 1 t, z 5 1, tPR.
b. First step is to use the first equation to remove x
from the second and third.
1 22x 2 3y 1 z 5 211
2 x 1 2y 1 z 5 2
3 2x 2 y 1 3z 5 212
So we have
4 0x 1 1y 1 3z 5 27, 1 1 2 3 2
5 0x 2 1y 2 5z 5 13, 1 2 2 3 3
Now the fourth and fifth equations are used to
create a sixth equation where the coefficient of
y is zero.
6 0x 1 0y 2 2z 5 6, 4 1 5
So 22z 5 6 or z 5 23.
9-41
Substituting this into equation 4 yields,
y 1 3(23) 5 27 or y 5 2. Finally substitute z and
y values into equation 2 to obtain the x value.
x 1 2(2) 1 (23) 5 2 or x 5 1.
Hence the solution to these three equations is
(1, 2, 23).
c. First step is to notice that the second equation is
simply twice the first equation.
1 2x 2 y 1 z 5 21
2 4x 2 2y 1 2z 5 22
3 2x 1 y 2 z 5 5
So the solution to these equations is the same as the
solution to just the first and third equations.
Moreover since this is two equations with three
unknowns, there will be infinitely many solutions.
4 4x 1 0y 1 0z 5 4, 1 1 3
Hence 4x 5 4 or x 5 1. Let y 5 t and solve for z
using the first equation.
2(1) 2 t 1 z 5 21, so z 5 23 1 t
Hence the solution to these equations is x 5 1,
y 5 t, z 5 23 1 t, tPR.
d. First step is to notice that the second equations
is simply twice the first and the third equation is
simply 24 times the first equation.
1 x 2 y 2 3z 5 1
2 2x 2 2y 2 6z 5 2
3 24x 1 4y 1 12z 5 24
So the solution to these equations is the same as the
solution to just the first equation. So the solution to
these equations is a plane. To solve this in parametric
equation form, simply let y 5 t and z 5 s and find
the x value.
x 2 t 2 3s 5 1, or x 5 1 1 t 1 3s
So the solution to these equations is x 5 1 1 3s 1 t,
y 5 t, z 5 s, s, tPR.
26. a. Since the normal of the first equation
is (1, 21, 1) and the normal of the second is
(1, 2, 22), which are not scalar multiples of each
other, there is a line of intersection between the
planes. The next step is to use the first and second
equations to find an equation with a zero for the
coefficient of x. The second equation minus the first
equation yields 0x 1 3y 2 3z 1 3 5 0. We may
divide by three to simplify, so y 2 z 1 1 5 0. If
we let z 5 t, then y 2 t 1 1 5 0, or y 5 21 1 t.
Substituting these into the first equation yields
x 2 (21 1 t) 1 t 2 1 5 0 or x 5 0. So the
equation of the line in parametric form is x 5 0,
y 5 21 1 t, z 5 t, tPR.
9-42
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 1 5 (0) 2 (21 1 t) 1 (t) 2 1
50
and
x 1 2y 2 2z 1 2 5 (0) 1 2(21 1 t) 2 2(t) 1 2
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The normal vector for the first plane is
(1, 24, 7), while the normal vector for the second
plane is (2, 28, 14) 5 2(1, 24, 7). Hence the
planes have collinear normal vectors, and so are
parallel.
The second equation is equivalent to
x 2 4y 1 7z 5 30, since we may divide the equation
by two. Since the constant on the right in the first
equation is 28, while the constant on the right in the
second equivalent equation is 30, these planes are
parallel and not coincident. So there is no intersection.
c. The normal vector for the first equation is
(1, 21, 1), while the normal vector for the second
equation is (2, 1, 1). Since the normal vectors are
not scalar multiples of each other, there is a line of
intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The second equation minus twice the first equation
yields 0x 1 3y 2 z 1 0 5 0.
Solving for z yields, z 5 3y. If we let y 5 t, then
z 5 3(t) 5 3t.
Substituting these into the first equation yields
x 2 (t) 1 (3t) 2 2 5 0 or x 5 2 2 2t. So the
equation of the line in parametric form is x 5 2 2 2t,
y 5 t, z 5 3t, tPR.
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 2 5 (2 2 2t) 2 (t) 1 (3t) 2 2
50
and
2x 1 y 1 z 2 4 5 2(2 2 2t) 1 (t) 1 (3t) 2 4
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
27. The angle, u, between the plane and the line is
the complementary angle of the angle between the
direction vector of the line and the normal vector
for the plane. The direction vector of the line is
Chapter 9: Relationships Between Points, Lines, and Planes
(1, 21, 0) and the normal vector for the plane is
(2, 0, 22).
0 (1, 21, 0) 0 5 "12 1 (21)2 1 02
5 "2
0 (2, 0, 22) 0 5 "22 1 02 1 (22)2 5 "8
(1, 21, 0) ? (2, 0, 22) 5 1(2) 2 1(0) 1 0(22) 5 2
So the angle between the normal vector and the
direction vector is cos21 A !22!8 B 5 60°. So
u 5 90 260° 5 30°.
> >
a?b
28. a. We have that cos (60°) 5 > > . Also
0a0 0b0 >
>
>
>
since a and b are unit vectors, 0 a 0 5 0 b 0 5 1 and
>
>
> >
a ? a 5 b ? b> 5 1, and moreover cos (60°) 5 12. So
>
> >
a?b
a?b5
5 12.
131
The dot >product is distributive,
so >
>
>
>
>
>
(6a 1 b ) ? (a 2 2b ) 5 6a ? (a 2 2b )
>
>
>
1 b ? (a 2 2b )
>
> >
>
5 6a ? a 1 6a ? (22b )
>
>
>
>
1 b ? a 1 b ? (22b )
> >
> >
> >
5 6a ? a> 2> 12a ? b 1 a ? b
2 2b ? b
1
1
5 6(1) 2 12a b 1 a b
2
2
2 2(1)
3
52
2 > >
x?y
b. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since
>
>
0 x 0 5 3, 0 y 0 5 4, and cos (60°) 5 12,
> >
> >
>
x ? y 5 12 (4)(3) 5 6. Also x ? x 5 0 x 0 2 5 9
> >
>2
and y ? y 5 0 y 0 5 16.
The dot product is distributive, so
>
>
>
>
>
>
>
(4x 2 y ) ? (2x 1 3y ) 5 4x ? (2x 1 3y )
>
>
>
2 y ? (2x 1 3y )
> >
> >
> >
5 8x ? x 1 12x ? y 2 2y ? x
> >
2 3y ? y
5 8(9) 1 12(6) 2 2(6)
2 3(16)
5 84
29. The origin, (0, 0, 0), and (21, 3, 1) are two
points on this line. So (21, 3, 1) is a direction vector
for this line and since the origin is on the line, a
>
possible vector equation is r 5 t(21, 3, 1), tPR.
(21, 3, 1) is a normal vector for the plane. So the
equation of the plane is 2x 1 3y 1 z 1 D 5 0.
Calculus and Vectors Solutions Manual
(21, 3, 1) is a point on the plane. Substitute the
coordinates to determine the value of D.
119111D50
D 5 211
The equation of the plane is 2x 1 3y 1 z 2 11 5 0.
30. The plane is the right bisector joining
P(21, 0, 1) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (0, 1, 21). So the line connecting the
two points is (21, 0, 1) 1 t(0, 1, 21), tPR, or in
corresponding
parametric form is x 5 21, y 5 t, z 5 1 2 t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we plug the parametric equation into the plane
equation and solve for t.
0x 1 y 2 z 5 0(21) 1 (t) 2 (1 2 t)
5 21 1 2t
So if y 2 z 5 0, then 21 1 2t 5 0, or t 5 12.
So the point of intersection is occurs at t 5 12, since
the origin point is P and the intersection occurs at
the midpoint of the line connecting P and its image,
the image point occurs at t 5 2 3 12 5 1. So the
image point is at x 5 21, y 5 1, z 5 1 2 (1) 5 0.
So the image point is (21, 1, 0).
31. a. Thinking of the motorboat’s velocity vector
(without the influence of the current) as starting
at the origin and pointing northward toward the
opposite side of the river, the motorboat has velocity
vector (0, 10) and the river current has velocity
vector (4, 0). So the resultant velocity vector of the
motorboat is
(0, 10) 1 (4, 0) 5 (4, 10)
To reach the other side of the river, the motorboat
needs to cover a vertical distance of 2 km. So the
hypotenuse of the right triangle formed by the
marina, the motorboat’s initial position, and the
motorboat’s arrival point on the opposite side of
the river is represented by the vector
1
4
(4, 10) 5 a , 2b
5
5
(We multiplied by 15 to create a vertical component
of 2 in the motorboat’s resultant velocity vector,
the distance needed to cross the river.) Since this
new vector has horizontal component equal to 45,
this means that the motorboat arrives 45 5 0.8 km
downstream from the marina.
9-43
b. The motorboat is travelling at 10 km> h, and in
part a. we found that it will travel along the vector
( 45, 2) . The length of this vector is
cos u 5
4
4 2
` a , 2b ` 5 a b 1 22
5
Å 5
5 "4.64
So the motorboat travels a total of !4.64 km to
cross the river which, at 10 km> h, takes
"4.64 4 10 8 0.2 hours
5 12 minutes.
32. a. Answers may vary. For example:
A direction
vector for this line is
>
AB 5 (6, 3, 4) 2 (2, 21, 3)
5 (4, 4, 1)
So, since the point B(6, 3, 4) is on this line, the
vector equation of this line is
>
r 5 (6, 3, 4) 1 t(4, 4, 1), tPR.
The equivalent parametric form is
x 5 6 1 4t
y 5 3 1 4t
z 5 4 1 t, tPR.
b. The line found in part a. will lie in the plane
x 2 2y 1 4z 2 16 5 0 if and only if both points
A(2, 21, 3) and B(6, 3, 4) lie in this plane.
We verify this by substituting these points into the
equation of the plane, and checking for consistency.
For A:
2 2 2(21) 1 4(3) 2 16 5 0
For B:
6 2 2(3) 1 4(4) 2 16 5 0
Since both points lie on the plane, so does the line
found in part a.
33. The wind velocity vector is represented by (16, 0),
and the water current velocity vector is represented
by (0, 12). So the resultant of these two vectors is
(16, 0) 1 (0, 12) 5 (16, 12).
Thinking of this vector with tail at the origin and
head at point (16, 12), this vector forms a right
triangle with vertices at points (0, 0), (0, 12), and
(16, 12). Notice that
0 (16, 12) 0 5 "162 1 122
5 "400
5 20
This means that the sailboat is moving at a speed
of 20 km> h once we account for wind and water
velocities. Also the angle, u, this resultant vector
makes with the positive y-axis satisfies
9-44
12
20
u 5 cos 21 a
12
b
20
8 53.1°
So the sailboat is travelling in the direction
N 53.1° E, or equivalently E 36.9° N.
34. Think of the weight vector for the crane with tail
at the origin at head at (0, 2400) (we use one unit
for every kilogram of mass). We need to express this
weight vector as the sum of two vectors: one that is
parallel to the inclined plane and pointing down this
>
incline (call this vector x 5 (a, b)), and one that is
perpendicular to the inclined plane and pointing
>
toward the plane (call this vector y 5 (c, d)). The
>
angle between x and (0, 2400) is 60° and the angle
>
>
>
between y and (0, 2400) is 30°. Of course, x and y
are perpendicular. Using the formula for dot product,
we get
>
>
y ? (0, 2400) 5 0 y 0 0 (0, 2400) 0cos 30°
2400d 5 400a
"3
b"c 2 1 d 2
2
22d 5 "3 ? "c 2 1 d 2
4d 2 5 3(c 2 1 d 2 )
d 2 5 3c 2
So, since c is positive and d is negative (thinking of
the inclined plane as moving upward from left to
>
right as we look at it means that y points down and
d
to the right), this last equation means that c 5 2"3
>
So a vector in the same direction as y is (1, 2"3).
>
We can find the length of y by computing the scalar
projection of (0, 2400) on (1, 2 !3), which equals
(0, 2400) ? (1, 2"3)
400"3
5
2
0 (1, 2"3) 0
5 200"3
>
That is, 0 y 0 5 200"3. Now we can find the length
>
of x as well by using the fact that
>
>
0 x 0 2 1 0 y 0 2 5 0 (0, 2400) 0 2
>
0 x 0 2 1 (200"3)2 5 4002
>
0 x 0 5 "160 000 2 120 000
5 "40 000
5 200
Chapter 9: Relationships Between Points, Lines, and Planes
So we get that
>
>
0 x 0 5 200 and 0 y 0 5 200"3. This means that the
component of the weight of the mass parallel to the
inclined plane is
>
9.8 3 0 x 0 5 9.8 3 200
5 1960 N,
and the component of the weight of the mass
perpendicular to the inclined plane is
>
9.8 3 0 y 0 5 9.8 3 200"3
8 3394.82 N.
35. a. True; all non-parallel pairs of lines intersect
in exactly one point in R2. However, this is not
the case for lines in R3 (skew lines provide a
counterexample).
b. True; all non-parallel pairs of planes intersect in a
line in R3.
c. True; the line x 5 y 5 z has direction vector
(1, 1, 1), which is not perpendicular to the normal
vector (1, 22, 2) to the plane x 2 2y 1 2z 5 k,
k any constant. Since these vectors are not
perpendicular, the line is not parallel to the plane,
and so they will intersect in exactly one point.
d. False; a direction vector for the line
z11
x
5 y 2 1 5 2 is (2, 1, 2). A direction vector
2
x21
y21
z11
for the line 24 5 22 5 22 is (24, 22, 22),
or (2, 1, 1) (which is parallel to (24, 22, 22)).
Since (2, 1, 2) and (2, 1, 1) are obviously not
parallel, these two lines are not parallel.
36. a. A direction vector for
y22
L1: x 5 2,
5z
3
is (0, 3, 1), and a direction vector for
z 1 14
L2: x 5 y 1 k 5
k
is (1, 1, k). But (0, 3, 1) is not a nonzero scalar
multiple of (1, 1, k) for any k since the first
Calculus and Vectors Solutions Manual
component of (0, 3, 1) is 0. This means that the
direction vectors for L1 and L2 are never parallel,
which means that these lines are never parallel for
any k.
b. If L1 and L2 intersect, in particular their
x-coordinates will be equal at this intersection point.
But x 5 2 always in L1 so we get the equation
25y1k
y522k
y22
Also, from L1 we know that z 5 3 , so substituting
this in for z in L2 we get
2k 5 z 1 14
y22
2k 5
1 14
3
3(2k 2 14) 5 y 2 2
y 5 6k 2 40
So since we already know that y 5 2 2 k, we
now get
2 2 k 5 6k 2 40
7k 5 42
k56
So these two lines intersect when k 5 6. We have
already found that x 5 2 at this intersection point,
but now we know that
y 5 6k 2 40
5 6(6) 2 40
5 24
y22
z5
3
24 2 2
5
3
5 22
So the point of intersection of these two lines is
(2, 24, 22), and this occurs when k 5 6.
9-45
CHAPTER 9
Relationships Between Points,
Lines, and Planes
Review of Prerequisite Skills, p. 487
1. a. Yes; (2, 25) 5 (10, 212) 1 t(8, 27)
(2, 25) 5 (10, 212) 1 1(8, 27)
b. No; 12(1) 1 5(2) 2 13 5 9 2 0
c. Yes; (7, 23, 8) 5 (1, 0, 24) 1 t(2, 21, 4)
(7, 23, 8) 5 (1, 0, 24) 1 3(2, 21, 4)
d. No; (1, 0, 5) 5 (2, 1, 22) 1 t(4, 21, 2)
(21, 21, 7) 2 t(4, 21, 2)
There is no value of t that satisfies the equation.
2. Answers may vary. For example:
>
a. Vector: m 5 (7, 3) 2 (2, 5) 5 (5, 22)
>
r 5 (2, 5) 1 t(5, 22), tPR
Parametric: x 5 2 1 5t, y 5 5 2 2t, tPR
>
b. Vector: m 5 (4, 27) 2 (23, 7) 5 (7, 214)
>
r 5 (23, 7) 1 t(7, 214), tPR
Parametric: x 5 23 1 7t, y 5 7 2 14t, tPR
>
c. Vector: m 5 (23, 211) 2 (21, 0)
5 (22, 211)
>
r 5 (21, 0) 1 t(22, 211), tPR
Parametric: x 5 21 1 2 2t, y 5 211t, tPR
>
d. Vector: m 5 (6, 27, 0) 2 (1, 3, 5)
5 (5, 210, 25)
>
r 5 (1, 3, 5) 1 t(5, 210, 25), tPR
Parametric: x 5 1 1 5t, y 5 3 2 10t, z 5 5 2 5t,
tPR
>
e. Vector: m 5 (21, 5, 2) 2 (2, 0, 21)
5 (23, 5, 3)
>
r 5 (2, 0, 21) 1 t(23, 5, 3), tPR
Parametric: x 5 2 2 3t, y 5 25t, z 5 21 1 3t,
tPR
>
f. Vector: m 5 (12, 25, 27) 2 (2, 5, 21)
5 (10, 210, 26)
>
r 5 (2, 5, 21) 1 t(10, 210, 26), tPR
Parametric: x 5 2 1 10t, y 5 5 2 10t, z 5 21 2 6t,
tPR
>
3. a. Since n 5 (2, 6, 21), the Cartesian equation
of the plane is of the form 2x 1 6y 2 z 1 D 5 0,
where D is to be determined. Since P0 (4, 1, 23)
is on the plane, it must satisfy the equation. So
Calculus and Vectors Solutions Manual
2(4) 1 6(1) 2 1(23) 1 D 5 8 1 6 1 3 1 D
5 17 1 D 5 0. D 5 217, and the equation of the
plane is 2x 1 6y 2 z 2 17 5 0.
>
b. Since n 5 (0, 7, 0), the Cartesian equation of
the plane is of the form 7y 1 D 5 0, where D is to
be determined. Since P0 (22, 0, 5) is on the plane, it
must satisfy the equation. So 7(0) 1 D 5 0 1 D 5 0
thus D 5 0. The equation of the plane is
7y 5 0, or y 5 0.
>
c. Since n 5 (4, 23, 0), the Cartesian equation of
the plane is of the form 4x 2 3y 1 D 5 0, where
D is to be determined. Since P0 (3, 21, 22)
is on the plane, it must satisfy the equation. So
4(3) 2 3(21) 1 D 5 12 1 3 1 D 5 15 1 D 5 0.
D 5 215, and the equation of the plane is
4x 1 3y 2 15 5 0.
>
d. Since n 5 (6, 5, 23), the Cartesian equation of
the plane is of the form 6x 2 5y 1 3z 1 D 5 0,
where D is to be determined. Since P0 (0, 0, 0) is on
the plane, it must satisfy the equation. So
6(0) 2 5(0) 1 3(0) 1 D 5 0, or D 5 0. The
equation of the plane is 6x 2 5y 1 3z 5 0.
>
e. Since n 5 (11, 26, 0), the Cartesian equation
of the plane is of the form 11x 2 6y 1 D 5 0,
where D is to be determined. Since P0 (4, 1, 8)
is on the plane, it must satisfy the equation. So
11(4) 2 6(1) 1 D 5 44 2 6 1 D 5 38 1 D 5 0.
D 5 238, and the equation of the plane is
11x 2 6y 2 38 5 0.
>
f. Since n 5 (1, 1, 21), the Cartesian equation of
the plane is of the form x 1 y 1 2 z 1 D 5 0,
where D is to be determined. Since P0 (2, 5, 1)
is on the plane, it must satisfy the equation.
So 2 1 5 2 1 1 D 5 6 1 D 5 0. D 5 26, and the
equation of the plane is x 1 y 2 z 2 6 5 0.
4. Start by writing the given line in parametric
form: (x, y, z) 5 (2 1 s 1 2t, 1 2 s, 3s 2 5t), so
x 5 2 1 s 1 2t, y 5 1 2 s, and z 5 3s 2 5t.
Solving for s in each component, we get s 5 1 2 y
and substituting this into z 5 3s 2 5t gives
z 5 3(1 2 y) 2 5t 5 3 2 3y 2 5t.
9-1
>
3 2 3y 2 z
.
t5
5
So now 23 1 3y 1 z 5 25t and
Finally, substituting both equations for s and t into
x 5 2 1 s 1 2t, we get
x 5 2 1 (1 2 y) 1 2 a
3 2 3y 2 z
b.
5
>
Rearranging, we get
5x 5 10 1 5 2 5y 1 6 2 6y 2 2z
5x 1 11y 1 2z 2 21 5 0.
5. L1 is not parallel to the plane because (3, 0, 2) is
a point on the line and the plane. Substitute the
expressions for the components of the parametric
equation of the line into the equation of the plane.
4(3 1 t) 1 (22t) 2 (2 1 2t) 2 10 5 0
12 1 4t 2 2t 2 2 2 2t 2 10 5 0
050
This last statement is always true. So every point on
the line is also in the plane. Therefore, the line lies
on the plane.
For L2 substitute the expressions for the components
of the parametric equation of the line into the
equation of the plane.
4(23t) 1 (25 1 2t) 2 (210t) 2 10 5 0
212t 2 5 1 2t 1 10t 2 10 5 0
215 5 0
This last statement is never true. So the line and the
plane have no points in common. Therefore, L2 is
parallel to the plane. The line cannot lie on the
plane.
For L3 use the symmetric equation to rewrite x and
z in terms of y.
x 5 24y 2 23
z 5 2y 2 6
Substitute into the equation of the plane.
4(24y 2 23) 1 y 2 (2y 2 6) 2 10 5 0
216y 2 92 1 y 1 y 1 6 2 10 5 0
214y 2 96 5 0
This equation has a solution. Therefore, L3 and the
plane have a point in common and are not parallel.
However, (5, 27, 1) is a point that lies on the line
that does not lie on the plane. Therefore, L3 does
not lie in the plane.
6. a. A normal vector to this plane is determined by
calculating
the
cross product of the position vectors,
>
>
AB and AC
.
>
AB> 5 (2, 0, 0) 2 (1, 0, 21) 5 (1, 0, 1)
AC 5 (6, 21, 5) 2 (1, 0, 21) 5 (5, 21, 6)
9-2
>
AB 3 AC 5 ((0 ? 6) 2 (1 ? 21), (1 ? 5)
2 (1 ? 6), (1 ? 21) 2 (0 ? 5))
5 (0 1 1, 5 2 6, 21 2 0)
>
5 (1, 21, 21) 5 n .
If P(x, y, z) is any point on the plane, then
AP 5 (x 2 1, y, z 1 1), and if the normal to the
plane is (1, 21, 21), then
(x 2 1, y, z 1 1) ? (1, 21, 21) 5 0, so
x 2 1 2 y 2 z 2 1 5 0 and thus,
x 2 y> 2 z 2 2 5 0
b. PQ> 5 (6, 4, 0) 2 (4, 1, 22) 5 (2, 3, 2)
PR 5 (0, >0, 23)> 2 (4, 1, 22) 5 (24, 21, 21)
>
n 5 PQ 3 PR
5 (3(21) 2 2(21)), 2(24) 22(21),
2(21) 2 3(4))
5 (23 1 2, 28 1 2, 22 1 12) 5 (21, 26, 10)
Since (21, 26, 10) 5 21(1, 6, 210), we will use
(1, 6, 210) as the normal vector so that the coefficient
of x is positive.
If P(x, y, z) is any point on the plane,
>
then AP 5 (x 2 4, y 2 1, z 1 2), and if the normal
to the plane is (1, 6, 210), then
(x 2 4, y 2 1, z 1 2) ? (1, 6, 210) 5 0,
so x 2 4 1 6y 2 6 2 10z 2 20 5 0,
and thus x 1 6y 2 10z 2 30 5 0.
7. Answers may vary. For example: One direction
>
vector is m 5 (2, 21, 6) 2 (1, 24, 3) 5 (1, 3, 3).
Now we need to find a normal to the plane such
>
>
that n ? m 5 0. So (1, 3, 3) ? (a, 0, c) 5 0. Now we
have a 1 3c 5 0. A possible solution to this is
>
a 5 3, c 5 21. So n 5 (3, 0, 21) and the
Cartesian equation of the plane is 3x 2 z 5 0.
Since the plane is parallel to the y-axis, (0, 1, 0) is
another direction vector for the plane. Therefore, a
vector equation for the plane is
>
r 5 (1, 24, 3) 1 t(1, 3, 3) 1 s(0, 1, 0), s, tPR.
8. We are given the point A(21, 3, 4). We need to
>
find a normal vector n 5 (a, b, c) such that
a(x 1 1) 1 b(y 2 3) 1 c(z 2 4) 1 d 5 0.
The normal vector also must be perpendicular to the
two planes and their normals, (2, 21, 3) and
(5, 1, 23). One possible solution for the normal is
>
n 5 (0, 3, 1). So we have
3(y 2 3) 1 z 2 4 5 0
3y 1 z 2 9 2 4 5 0
And the equation of the plane is 3y 1 z 5 13.
Chapter 9: Relationships Between Points, Lines, and Planes
9.1 The Intersection of a Line
with a Plane and the Intersection
of Two Lines, pp. 496–498
1. a. First, show the parametric equations as
x 5 1 1 5s, y 5 2 1 s, z 5 23 1 s. Then the
plane can be written as p: x 2 2y 2 3z 5 6,
and the vector equation of the line is
>
r 5 (1, 2, 23) 1 s(5, 1, 1), sPR.
b. When we substitute the parametric equations into
the Cartesian equation for the plane, we get
(1 1 5s) 2 2(2 1 s) 2 3(23 1 s) 5 6
1 2 4 1 9 1 5s 2 2s 2 3s 5 6 2 0s 5 6
Note that by finishing the solution, we get 0s 5 0.
Since any real number will satisfy this equation, we
have an infinite number of solutions, and this line
lies on the plane.
2. a. A line and a plane can intersect in three ways:
Case 1: The line and the plane have zero points of
intersection. This occurs when the lines are not
incidental, meaning they do not intersect.
Case 2: The line and the plane have only one point
of intersection. This occurs when the line crosses
the plane at a single point.
Case 3: The line and the plane have an infinite
number of intersections. This occurs when the line
is coincident with the plane, meaning the line lies
on the plane.
b. Assume that the line and the plane have more
than one intersection, but not an infinite number.
For simplicity, assume two intersections. At the
first intersection, the line crosses the plane. In order
for the line to continue on, it must have the same
direction vector. If the line has already crossed the
plane, then it continues to move away from the
plane, and can not intersect again. So the line and
the plane can only intersect zero, one, or infinitely
many times.
>
3. a. The line r 5 s(1, 0, 0) is the x-axis.
>
b. The plane y 5 1 has the form r 5 (x, 1, z),
where x, and z are any values in R. So the plane is
parallel to the xz-plane, but just one unit away to
the right.
Calculus and Vectors Solutions Manual
c.
z
x
y
d. There are no intersections between the line and
the plane.
4. a. For x 1 4y 1 z 2 4 5 0, if we substitute the
parametric equations, we have
(22 1 t) 1 4(1 2 t)1 (2 1 3t) 1 4
5 22 1 4 1 2 1 t 2 4t 1 3t 2 4
5 0t 1 0
5 0. All values of t give a solution to the equation,
so all points on the line are also on the plane.
b. For the plane 2x 2 3y 1 4z 2 11 5 0, we can
substitute the parametric equations derived
>
from r 5 (1, 5, 6) 1 t(1, 22, 22):
x 5 1 1 t, y 5 5 2 2t, z 5 6 2 2t.
So we have 2(1 1 t)2 3(5 2 2t) 1 4(6 2 2t) 2 11
5 2 2 15 1 24 2 11 1 2t 1 6t 2 8t
5 0t 1 0
50
Similar to part a., all values of t give a solution to
this equation, so all points on the line are also on
the plane.
5. a. First, we should determine the parametric
equations from the vector form: x 5 21 2 s,
y 5 1 1 2s, z 5 2s. Substituting these into the
equation of the plane, we get
2(21 2 s) 2 2(1 1 2s) 1 3(2s) 2 1
5 22 2 2 2 1 2 2s 2 4s 1 6s
5 25 1 0s
Since there are no values of s such that 25 5 0, this
line and plane do not intersect.
b. Substituting the parametric equations into the
equation of the plane, we get
2(1 1 2t) 2 4(22 1 5t) 1 4(1 1 4t) 2 13
2 1 8 1 4 2 13 1 4t 2 20t 1 16t
5 1 1 0t
Since there are no values of t such that 1 5 0, there
are no solutions, and the plane and the line do not
intersect.
9-3
>
6. a. The direction vector is m 5 (21, 2, 2)
>
and the normal is n 5 (2, 22, 3), so if the line and
> >
the plane meet at right angles, m ? n 5 0. So
(21 ? 2) 1 (2 ? 22) 1 (2 ? 3) 5 22 2 4 1 6 5 0,
but 2(21) 2 2(1) 1 3(0) 2 1 5 25 2 0. So the
point on the line is not on the plane.
>
b. The direction vector is m 5 (2, 5, 4) and
>
the normal is n 5 (2, 24, 4), so if the line
> >
and the plane meet at right angles, m ? n 5 0. So
(2 ? 2) 1 (5 ? 24) 1 (4 ? 4) 5 4 2 20 1 16 5 0,
but 2(1) 2 4(22) 1 4(1) 2 13 5 1 2 0. So the
point on the line is not on the plane.
7. a. If the line and the plane intersect, then they are
equal at a particular point p. So we must substitute
the parametric equations into the equation of the
plane, and then solve for p.
(21 1 6p) 1 2(3 1 p) 2 (4 2 2p) 1 29
5 21 1 6 2 4 1 6p 1 2p 1 2p 1 29
5 30 1 10p
5 0. So now 210p 5 30 and p 5 23.
Now we must find the point at which the
line and plane intersect. To do this, just
substitute p 5 23 into the vector form of the line:
(21, 3, 4) 1 23(6, 1, 22) 5 (219, 0, 10).
b. If the line and the plane intersect, then they are
equal at a particular point p. So we must substitute
the parametric equations into the equation of the
plane, and then solve for p.
x 5 1 1 4s, y 5 22 2 s, z 5 3 1 s
2(1 1 4s) 1 7(22 2 s) 1 (3 1 s) 1 15
5 2 2 14 1 3 1 15 1 8s 2 7s 1 s
5 6 1 2s
5 0. So now 22s 5 6 and s 5 23. Now we must
find the point at which the line and plane intersect.
To do this, just substitute s 5 23 into the vector
form of the line:
(1, 22, 3) 1 23(4, 21, 1) 5 (211, 1, 0)
8. a. Comparing the x and y components in L1 and
L2, we have
3 1 4s 5 4 1 13t
1 2 s 5 1 2 5t
We can easily solve for one of the variables by
using the second equation: s 5 5t. Substituting this
back into the first equation: 3 1 20t 5 4 1 13t so
1 5 7t and thus t 5 17. So now we must solve for s:
5
3 1 4s 5 4 1 137 and s 5 20
28 5 7 . Placing these back
into the equations for L1 and L2:
9-4
5
41 2 45
L1: (3, 1, 5) 1 (4, 21, 2) 5 a , , b
7
7 7 7
13
5 5
41 2 5
L2: a4 1 , 1 2 , b 5 a , , b
7
7 7
7 7 7
The points must be equal for intersection to occur,
so there is no intersection and the lines are skew.
b. If we compare the z components of the two lines,
we see 2 5 8 2 6s or s 5 1. Substituting this back
into the x component (the y component would work
just as well), we have 3 1 m 5 23 1 7(1) 5 4, or
m 5 1. So now we can substitute m and s back into
the equations for the line, and we get
L3 : (3, 7, 2) 1 (1, 26, 0) 5 (4, 1, 2)
L4 : (23, 2, 8) 1 (7, 21, 26) 5 (4, 1, 2)
So (4, 1, 2) is the only point of intersection between
these two lines.
9. a. Comparing the y and z components of each
vector equation, we get the system of equations:
3 2 2p 5 3 2 2q
4 1 3p 5 24 1 11q
Note that from the first equation, p 5 q. So the
second equation becomes 4 1 3q 5 24 1 11q.
Solving for q, we get q 5 1. So from the earlier
relation, p 5 1. Placing these two values back into
the vector equations, we get
(22, 3, 4) 1 (6, 22, 3) 5 (4, 1, 7)
(22, 3, 24) 1 (6, 22, 11) 5 (4, 1, 7)
This shows that these two lines intersect at (4, 1, 7).
b. Comparing the x and z components of each
vector equation, we get the system of equations:
41r521s
6 1 4r 5 28 1 5s
Note that from the first equation, s 5 2 1 r. So the
second equation becomes 6 1 4r 5 2 1 5r.
Solving for r, we get r 5 4. So from the earlier
relation, s 5 6. Placing these two values back into
the vector equations, we get
(4, 1, 6) 1 4(1, 0, 4) 5 (8, 1, 22)
(2, 1, 28) 1 6(1, 0, 5) 5 (8, 1, 22)
This shows that these two lines intersect
at (8, 1, 22).
c. Comparing the x and z components of each
vector equation, we get the system of equations:
2 1 m 5 22 1 3p
11m512p
Note that from the second equation, m 5 2p. So
the first equation becomes 2 2 p 5 22 1 3p.
Solving for p, we get p 5 1. So from the earlier
relation, m 5 21. Placing these two values back
into the vector equations, we get
Chapter 9: Relationships Between Points, Lines, and Planes
(2, 2, 1) 2 (1, 1, 1) 5 (1, 1, 0)
(22, 2, 1) 1 (3, 21, 21) 5 (1, 1, 0)
This shows that these two lines intersect at (1, 1, 0).
d. Comparing the x and y components of each
vector equation, we get the system of equations:
1 1 0m 5 2 1 s
2 1 4m 5 3 2 2s
Note that from the first equation, s 5 21. So the
second equation becomes 2 1 4m 5 5.
Solving for m, we get m 5 34. Placing these two
values back into the vector equations, we get
3
21
(9, 1, 2) 2 (5, 0, 4) 5 a , 1, 21b
4
4
(8, 2, 3) 2 (4, 1, 22) 5 (4, 1, 5)
The two lines do not intersect, so they are skew.
10. At the point where the line intersects the z-axis,
the point Q(0, 0, q) equals the vector equation. So
for the x component, 23 1 3s 5 0 or s 5 1.
Substituting this into the vector equation, we get
(23, 2, 1) 1 (3, 22, 7) 5 (0, 0, 8). So q 5 8.
11. a. Comparing the x components, we get
22 1 7s 5 230 1 7t, which can be reduced to
28 1 7s 5 7t or s 2 t 5 4. Comparing the other
components, the same equation results.
b. From L1, we see that at (22, 3, 4), s 5 0. When
this occurs, t 5 4. Substituting this into L2, we get
(230, 11, 24) 1 4(7, 22, 2) 5 (22, 3, 4). Since
both of these lines have the same direction vector
and a common point, the lines are coincidental.
12. a. First, we must determine the values of s and t.
So comparing the x and z components, we get
23 1 s 5 1 2 3t
1 1 s 5 2 1 8t
From the second equation, s 5 1 1 8t. Substituting
this back into the first equation,
23 1 1 1 8t 5 1 2 3t or t 5 113 .
Substituting back into the second equation,
23 1 s 5 1 2 119 5 112 , and solving for s,
s 5 112 1 3 5 35
11 . Now we can solve for k. Compare
the y components after substituting s and t.
35
3
82
541 k
11
11
53 5 44 1 3k
or k 5 3.
b. The lines intersect when s 5 35
11 . The point of
35
35
intersection is ( 23 1 35
11 , 8 2 11 , 1 1 11 ) or
( 112 , 5311, 4611) .
Calculus and Vectors Solutions Manual
13. On the xz-plane, the point A has the coordinates
(x, 0, z), for any x, z. Similarly, on the yz-plane, the
point B has the coordinates (0, y, z) for any y, z. Now
the task is to find the required values of s for these
points. Starting with the x component of point B,
we have 0 5 28 1 2s or s 5 4. So point B is
(28, 26, 21) 1 4(2, 2, 1) 5 (0, 2, 3). For point A,
we need the y coordinate to equal 0. So 0 5 26 1 2s
or s 5 3. So point A is
(28, 26, 21) 1 3(2, 2, 1) 5 (22, 0, 2).
Now we need to find the distance.
d 5 "(0 2 (22))2 1 (2 2 0)2 1 (3 2 2)2
5 "4 1 4 1 1
5 "9
53
14. a. Comparing the y and z components of each
vector equation, we get the system of equations:
1 1 0p 5 21 2 2q
1 2 p 5 1 2 2q
Note that from the first equation, 2 5 22q or
q 5 21. So the second equation becomes
1 2 p 5 1 1 2 or p 5 22.
Placing these two values back into the vector
equations to find the intersection point A, we get
(2, 1, 1) 2 2(4, 0, 21) 5 (26, 1, 3)
(3, 21, 1) 2 (9, 22, 22) 5 (26, 1, 3)
Thus, the intersection point is (26, 1, 3).
b. A point on the xy plane has the form (x, y, 0). If
such a point is (26, 1, 0) then the distance from
this point is d 5 "0 1 0 1 32 5 3.
15. a. Comparing the x and y components of each
vector equation, we get the system of equations:
21 1 5s 5 4 1 0t
3 2 2s 5 21 1 2t
Note that from the first equation, 5 5 5s or s 5 1.
So the second equation becomes 3 2 2 5 21 1 2t
or t 5 1. Placing these two values back into the
vector equations to find the intersection point A,
we get
(21, 3, 2) 1 (5, 22, 10) 5 (4, 1, 12)
(4, 21, 1) 1 (0, 2, 11) 5 (4, 1, 12)
Thus, the intersection point is (4, 1, 12).
b. We need to find a vector (a, b, c) such that
5a 2 2b 1 10c 5 0
2b 1 11c 5 0
A possible solution to the second equation is
(a, 11, 22). If we substitute this into the first
equation, we get 5a 2 22 2 20 5 0 S 5a 5 42.
9-5
We can use this to get a solution of ( 425, 11, 22) . To
eliminate the fraction, we get (42, 55, 210). So the
>
vector equation is r 5 (4, 1, 12) 1 t(42, 55, 210),
tPR.
16. a.
z
L2
x
L1
y
b. The only point of intersection is at the origin
(0, 0, 0).
c. If p 5 0 and q 5 0, the intersection occurs at
(0, 0, 0).
17. a. Represent the lines parametrically, and
then substitute into the equation for the plane.
For the first equation, x 5 t, y 5 7 2 8t,
z 5 1 1 2t. Substituting into the plane equation,
2t 1 7 2 8t 1 3 1 6t 2 10 5 0. Simplifying,
0t 5 0. So the line lies on the plane.
For the second line, x 5 4 1 3s, y 5 21, z 5 1 2 2s
Substituting into the plane equation,
8 1 6s 2 1 1 3 2 6s 2 10 5 0. Simplifying,
0s 5 0. This line also lies on the plane.
b. Compare the x and y components:
4 1 3s 5 t
7 2 8t 5 21
From the second equation, t 5 1. Substituting back
into the first equation, 4 1 3s 5 1, or s 5 21.
Determine the point of intersection:
(1, 7 2 8, 1 1 2) 5 (1, 21, 3)
(4 2 3, 21, 1 1 2) 5 (1, 21, 3)
The point of intersection is (1, 21, 3).
18. Answers may vary. For example:
>
r 5 (2, 0, 0) 1 p(2, 0, 1), pPR
9.2 Systems of Equations, pp. 507–509
1. a. linear
b. not linear
c. linear
d. not linear
2. Answers may vary. For example:
x 1 y 1 2z 5 215
a. x 1 2y 1 z 5 23
2x 1 y 1 z 5 210
b. Subtract the first equation from the second, and
subtract twice the first equation from the third.
9-6
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 2 y 2 3z 5 20
Add the second and third equation.
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 1 0y 2 4z 5 32
From the third equation, z 5 28.
Substitute z back into the second equation,
2y 2 8 5 212
2y 5 212 1 8 5 24
So y 5 4. Now substitute y and z back into the first
equation.
x 1 4 1 2(28) 5 x 2 12 5 215
And so x 5 23. Thus the solution is (23, 4, 28)
as expected.
3
3. a. 27 2 3(5) 1 4 a b 5 27 2 15 1 3 5 219
4
3
27 2 8 a b 5 27 2 6 5 213
4
27 1 2(5) 5 3
Yes, ( 27, 5, 34) is a solution.
b.
3
3(27) 2 2(5) 1 16 a b 5 221 2 10 1 12 5 219
4
3(27) 2 2(5) 5 221 2 10 5 231
2 223
3
8(27) 2 5 1 4 a b 5 256 2 5 1 3 5 258
4
Because the second equation fails to produce an
equality, ( 27, 5, 34) is not a solution.
4. a. Solve for y. y 5 23
The solution is (22, 23).
b. Multiply the second equation by 6
3x 1 5y 5 221
x 2 3y 5 7
Add 3 times the first equation to 5 times the second
equation.
3x 1 5y 5 221
14x 5 228
From the second equation, x 5 22.
Substituting x back into the first equation,
3(22) 1 5y 5 221
5y 5 215
So y 5 23.
The two systems are equivalent because they have
the same solution.
5. a. Add the second equation to 5 times the first
equation.
Chapter 9: Relationships Between Points, Lines, and Planes
We can use this to get a solution of ( 425, 11, 22) . To
eliminate the fraction, we get (42, 55, 210). So the
>
vector equation is r 5 (4, 1, 12) 1 t(42, 55, 210),
tPR.
16. a.
z
L2
x
L1
y
b. The only point of intersection is at the origin
(0, 0, 0).
c. If p 5 0 and q 5 0, the intersection occurs at
(0, 0, 0).
17. a. Represent the lines parametrically, and
then substitute into the equation for the plane.
For the first equation, x 5 t, y 5 7 2 8t,
z 5 1 1 2t. Substituting into the plane equation,
2t 1 7 2 8t 1 3 1 6t 2 10 5 0. Simplifying,
0t 5 0. So the line lies on the plane.
For the second line, x 5 4 1 3s, y 5 21, z 5 1 2 2s
Substituting into the plane equation,
8 1 6s 2 1 1 3 2 6s 2 10 5 0. Simplifying,
0s 5 0. This line also lies on the plane.
b. Compare the x and y components:
4 1 3s 5 t
7 2 8t 5 21
From the second equation, t 5 1. Substituting back
into the first equation, 4 1 3s 5 1, or s 5 21.
Determine the point of intersection:
(1, 7 2 8, 1 1 2) 5 (1, 21, 3)
(4 2 3, 21, 1 1 2) 5 (1, 21, 3)
The point of intersection is (1, 21, 3).
18. Answers may vary. For example:
>
r 5 (2, 0, 0) 1 p(2, 0, 1), pPR
9.2 Systems of Equations, pp. 507–509
1. a. linear
b. not linear
c. linear
d. not linear
2. Answers may vary. For example:
x 1 y 1 2z 5 215
a. x 1 2y 1 z 5 23
2x 1 y 1 z 5 210
b. Subtract the first equation from the second, and
subtract twice the first equation from the third.
9-6
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 2 y 2 3z 5 20
Add the second and third equation.
x 1 y 1 2z 5 215
0x 2 y 1 z 5 212
0x 1 0y 2 4z 5 32
From the third equation, z 5 28.
Substitute z back into the second equation,
2y 2 8 5 212
2y 5 212 1 8 5 24
So y 5 4. Now substitute y and z back into the first
equation.
x 1 4 1 2(28) 5 x 2 12 5 215
And so x 5 23. Thus the solution is (23, 4, 28)
as expected.
3
3. a. 27 2 3(5) 1 4 a b 5 27 2 15 1 3 5 219
4
3
27 2 8 a b 5 27 2 6 5 213
4
27 1 2(5) 5 3
Yes, ( 27, 5, 34) is a solution.
b.
3
3(27) 2 2(5) 1 16 a b 5 221 2 10 1 12 5 219
4
3(27) 2 2(5) 5 221 2 10 5 231
2 223
3
8(27) 2 5 1 4 a b 5 256 2 5 1 3 5 258
4
Because the second equation fails to produce an
equality, ( 27, 5, 34) is not a solution.
4. a. Solve for y. y 5 23
The solution is (22, 23).
b. Multiply the second equation by 6
3x 1 5y 5 221
x 2 3y 5 7
Add 3 times the first equation to 5 times the second
equation.
3x 1 5y 5 221
14x 5 228
From the second equation, x 5 22.
Substituting x back into the first equation,
3(22) 1 5y 5 221
5y 5 215
So y 5 23.
The two systems are equivalent because they have
the same solution.
5. a. Add the second equation to 5 times the first
equation.
Chapter 9: Relationships Between Points, Lines, and Planes
2x 2 y 5 11
11x 5 66
Solve for x in the second equation, x 5 6. Substitute x
back into the first equation
2(6) 2 y 5 11
2y 5 11 2 12 5 21
So y 5 1
Therefore, the solution is (6, 1).
b. Subtract three times the first equation from twice
the second equation.
2x 1 5y 5 19
0x 2 7y 5 235
From the second equation, y 5 5.
Substitute y back into the first equation.
2x 1 5(5) 5 19
2x 5 19 2 25 5 26
So x 5 23
Therefore, the solution is (23, 5).
c. Add the second equation to 3 times the first
equation to the second equation
2x 1 2y 5 10
0x 1 11y 5 33
From the second equation, y 5 3.
Substitute y back into the first equation.
2x 1 2(3) 5 10
2x 5 4
So x 5 24.
Therefore the solution is (24, 3).
6. a. These two lines are parallel, and therefore
cannot have an intersection.
b. The second equation is five times the first,
therefore the lines are coincident.
7. a. Let x 5 t. So 2t 2 y 5 3 then y 5 2t 2 3.
b. Let x 5 t, y 5 s. Then t 2 2s 1 z 5 0 and
z 5 2s 1 t.
8. a. If x 5 t, y 5 22t 2 11, then y 5 22x 2 11
and so 2x 1 y 5 211 is the required linear equation.
b. 2x 1 y 5 211
2(3t 1 3) 1 (26t 2 17) 5 6t 2 6t 1 6 2 17
5 211
9. a. The two equations will have no solutions when
k 2 12, since they will be parallel should this occur.
b. It is impossible to have only one solution for these
two equations. They have exactly the same direction
vector. They will never intersect at exactly one place.
c. The two equations will have infinitely many
solutions when k 5 12. When this occurs, the two
equations are coincident.
Calculus and Vectors Solutions Manual
10. a. There are infinitely many solutions to this
equation. This is reason why it is represented
graphically as a line.
b. Let x 5 t. So 2t 1 4y 5 11, then 4y 5 11 2 2t
and y 5 114 2 12 t, tPR
c. This equation will not have any integer solutions
because the left hand side is an even function and
the right side is an odd function.
11. a. Add the second equation to 22 times the
first.
x 1 3y 5 a
0x 2 3y 5 b 2 2a
Divide the second equation by 23 to get
y 5 2 13 b 1 23 a. Now substitute this back into
the first equation.
1
2
x 1 3a2 b 1 ab 5 a
3
3
x 2 b 1 2a 5 a
x 5 2a 1 b
b. Since they have different direction vectors, these
two equations are not parallel or coincident and will
intersect somewhere.
12. a. Add the third equation to the first to eliminate z.
x1y1z50
x 2 y 1 0z 5 1
x 1 2y 1 0z 5 25
Add twice the second equation to the third equation
to eliminate
Add twice the second equation to the third equation
to eliminate y.
x1y1z50
x 2 y 1 0z 5 1
3x 1 0y 1 0z 5 23
Divide the third equation by 23 to get x 5 21.
Now substitute into the second equation.
21 2 y 5 1
y 5 22
Finally, substitute x and y to get
21 1 22 1 z 5 0
So z 5 3. Therefore, the solution is (21, 22, 3).
b. Add the first equation to 22 times the second,
and add the first equation to 22 times the third.
2x 2 3y 1 z 5 6
0x 2 5y 2 3z 5 256
0x 2 y 1 3z 5 40
Now add the second equation to 21 times the third.
2x 2 3y 1 z 5 6
0x 2 5y 2 3z 5 256
0x 2 4y 1 0z 5 216
9-7
From the third equation, y 5 4.
Now substitute this into the second equation.
25(4) 2 3z 5 256
23z 5 236
z 5 12
Now substitute these two values back into the first
equation.
2x 2 3(4) 1 12 5 6
2x 5 6, x 5 3
So the solution is (3, 4, 12).
c. Add the second equation to 21 times the third.
x 1 y 1 0z 5 10
0x 1 y 1 z 5 22
2x 1 y 1 0z 5 2
Add the third equation to the first equation.
x 1 y 1 0z 5 10
0x 1 y 1 z 5 22
0x 1 2y 1 0z 5 12
So y 5 6. Now substitute into the other two
equations.
x 1 6 5 10 S x 5 4
6 1 z 5 22 S z 5 28
So the solution is (4, 6, 28).
d. To eliminate fractions, multiply each of the
equations by 60.
20x 1 15y 1 12z 5 840
15x 1 12y 1 20z 5 21260
12x 1 20y 1 15z 5 420
Add 3 times the first equation to 24 times the
second, and add 3 times the first equation to 25
times the third.
20x 1 15y 1 12z 5 840
0x 2 3y 2 44z 5 7560
0x 2 55y 2 39z 5 420
Now add 55 times the second equation to 23 times
the third equation.
20x 1 15y 1 12z 5 840
0x 2 3y 2 44z 5 7560
0x 1 0y 2 2303z 5 414 540
Divide the third equation through by 22303 to get
z 5 2180. Substituting z back into the second
equation.
23y 2 44(2180) 5 7560 S 23y 5 2360
So y 5 120. Now substitute these two values back
into the first equation.
20x 1 15(120) 1 12(2180) 5 840
20x 5 840 2 1800 1 2160 5 1200
So x 5 60. Therefore the solution is (60, 120, 2180).
9-8
e. Note that if 2x 2 y 5 0 S y 5 2x, and
2z 2 x 5 0 S z 5 12 x. So we substitute these two
relations into the second equation.
1
7
2(2x) 2 x 5 x 5 7 S x 5 2
2
2
So now z 5 1, y 5 4, and the solution is (2, 4, 1).
f. Add the first equation to 22 times the second
equation.
x 1 y 1 2z 5 13
22x 1 0y 2 7z 5 238
2x 1 0y 1 6z 5 32
Add the second and third equations.
x 1 y 1 2z 5 13
22x 1 0y 2 7z 5 238
0x 1 0y 2 z 5 26
So from the third equation, z 5 6.
Substituting into the second equation,
22x 2 42 5 238
22x 5 4 S x 5 22
Finally, substituting both values into the first
equation,
22 1 y 1 12 5 13 S y 5 3.
So the final solution is (22, 3, 6).
13. Answers may vary. For example:
a.
Three lines parallel
z
L3
L2
L1
y
x
Two lines coincident
and the third parallel
z
L3
L2
L1
y
x
Chapter 9: Relationships Between Points, Lines, and Planes
Two parallel lines cut
by the third line
z
L3
L2
L1
y
x
The lines form a triangle
z
L2
L3
b.
y
L1
x
Lines meet in a point
z
L1
L3
L2
y
x
c.
Three coincident lines
z
L1
L2
L3
y
x
14. a. Add 21 times the first equation and the
second equation. Add 21 times the first equation
and the third equation.
x1y1z5a
0x 1 0y 2 z 5 b 2 a
2x 1 0y 1 0z 5 c 2 a
Calculus and Vectors Solutions Manual
So z 5 a 2 b, x 5 a 2 c. Then substitute into the
first equation.
a2c1y1a2b5a
y 5 2a 1 b 1 c
So the final solution is (a 2 c, 2a 1 b 1 c, a 2 b).
15. a. For two equations to have no solutions, they
must be parallel—meaning it must have a the same
direction vector. So if k 5 2, then the lines are
parallel.
b. If two equations have an infinite number of
solutions, then the lines must be coincident. One
way to do this is if the second equation is a multiple
of the first equation. To achieve this, k 5 22.
c. For two equations to have a unique solution, they
must have only one intersection. From a., we saw
that if k 5 2, the lines are parallel, and if k 5 22,
then they are coincident. Since the only other option
is for the lines to have a unique solution, k 2 6 2.
9.3 The Intersection of Two Planes,
pp. 516–517
1. a. This means that the two equations represent
planes that are parallel and not coincident.
b. Answers may vary. For example: x 2 y 1 z 5 1,
x 2 y 1 z 5 22
2. a. The solution to the system of equations is:
x 5 12 1 12 s 2 t, y 5 s, z 5 t, s, tPR. The two
planes are coincident.
b. Answers may vary. For example:
x 2 y 1 z 1 21; 2x 2 2y 1 2z 5 22
3. a. 2z 5 24 1 z 5 22.
x 2 y 1 (22) 1 21
x 2 y 1 1.
x 5 1 1 s, y 5 s, z 5 22, sPR
The two planes intersect in a line.
b. Answers may vary. For example:
x 2 y 1 z 5 21; x 2 y 2 z 5 3.
4. a. 1 2x 1 y 1 6z 5 p; 2 x 1 my 1 3z 5 q
For the planes to be coincident equation 2 must be
a multiple of equation 1 . Since the coefficients of
x and z in equation 1 are twice that of the x and z
coefficients in equation 2 all of the coefficients
and constants in equation 2 must be half of the
corresponding coefficients in equation 1 . So:
1
m 5 , p 5 2q. q 5 1, and p 5 2.
2
The value for m is unique, but p just has to be twice
q and arbitrary values can be chosen.
9-9
Two parallel lines cut
by the third line
z
L3
L2
L1
y
x
The lines form a triangle
z
L2
L3
b.
y
L1
x
Lines meet in a point
z
L1
L3
L2
y
x
c.
Three coincident lines
z
L1
L2
L3
y
x
14. a. Add 21 times the first equation and the
second equation. Add 21 times the first equation
and the third equation.
x1y1z5a
0x 1 0y 2 z 5 b 2 a
2x 1 0y 1 0z 5 c 2 a
Calculus and Vectors Solutions Manual
So z 5 a 2 b, x 5 a 2 c. Then substitute into the
first equation.
a2c1y1a2b5a
y 5 2a 1 b 1 c
So the final solution is (a 2 c, 2a 1 b 1 c, a 2 b).
15. a. For two equations to have no solutions, they
must be parallel—meaning it must have a the same
direction vector. So if k 5 2, then the lines are
parallel.
b. If two equations have an infinite number of
solutions, then the lines must be coincident. One
way to do this is if the second equation is a multiple
of the first equation. To achieve this, k 5 22.
c. For two equations to have a unique solution, they
must have only one intersection. From a., we saw
that if k 5 2, the lines are parallel, and if k 5 22,
then they are coincident. Since the only other option
is for the lines to have a unique solution, k 2 6 2.
9.3 The Intersection of Two Planes,
pp. 516–517
1. a. This means that the two equations represent
planes that are parallel and not coincident.
b. Answers may vary. For example: x 2 y 1 z 5 1,
x 2 y 1 z 5 22
2. a. The solution to the system of equations is:
x 5 12 1 12 s 2 t, y 5 s, z 5 t, s, tPR. The two
planes are coincident.
b. Answers may vary. For example:
x 2 y 1 z 1 21; 2x 2 2y 1 2z 5 22
3. a. 2z 5 24 1 z 5 22.
x 2 y 1 (22) 1 21
x 2 y 1 1.
x 5 1 1 s, y 5 s, z 5 22, sPR
The two planes intersect in a line.
b. Answers may vary. For example:
x 2 y 1 z 5 21; x 2 y 2 z 5 3.
4. a. 1 2x 1 y 1 6z 5 p; 2 x 1 my 1 3z 5 q
For the planes to be coincident equation 2 must be
a multiple of equation 1 . Since the coefficients of
x and z in equation 1 are twice that of the x and z
coefficients in equation 2 all of the coefficients
and constants in equation 2 must be half of the
corresponding coefficients in equation 1 . So:
1
m 5 , p 5 2q. q 5 1, and p 5 2.
2
The value for m is unique, but p just has to be twice
q and arbitrary values can be chosen.
9-9
b. For parallel planes all of the coefficients of the
variables must be multiples of each other, but the
constant terms must differ by a different constant.
So a possible solution is:
1
m 5 , q 5 1, and p 5 3.
2
The value for m is again unique but p and q can be
arbitrarily chosen as long as p 2 2q.
c. For the two planes to intersect at right angles the
>
two normal vectors, n1 5 (2, 1, 6) and
>
n2 5 (1, m, 3), must satisfy:
>
>
n1 ? n2 5 0.
>
>
n1 ? n2 5 2 1 m 1 18 5 0
m 5 220. This value is unique, since only one
value was found to satisfy the given conditions.
d. From c. we know that in order to intersect in
right angles m 5 220. Choose p 5 1, q 5 1.
The value for m is unique from the solution to c.,
but the values for p and q can be arbitrary since the
only value which can change the angle between the
planes is m.
5. a. Letting z 5 s:
y 5 23s.
x 1 2(23s) 2 3s 5 0.
x 5 9s
The solution is:
x 5 9s, y 5 23s, z 5 s, sPR
b. Letting y 5 t.
t 1 3z 5 0
3z 5 2t
1
z 5 2 t.
3
1
x 1 2t 2 3a2 bt 5 0
3
x 1 3t 5 0
x 5 23t.
The solution is:
1
x 5 23t, y 5 t, z 5 2 t; tPR.
3
c. Since t is an arbitrary real number we can
express t as:
t 5 23s; sPR.
Substituting this into the solution for b. shows that
the two solutions are equivalent.
6. a. Equation 2 is twice that of equation 1 , so
they represent intersecting coincident planes.
b. The coefficients of each variable are the same,
but the constant terms are different, so the equations
represent non-intersecting parallel planes.
9-10
c. The coefficients of the x and z variables are the
same but the y coefficients are different. So the
equations represent planes that intersect in a line.
d. The coefficients of each variable from equation 1
to 2 are not the same multiple. Therefore the
equations represent planes that intersect in a line.
e. The intersection is a line by the same reasoning
as d.
f. The intersection is a line by the same reasoning as d.
7. a. x 5 1 2 s 2 t, y 5 s, z 5 t, s, tPR
b. There is no solution since the planes are parallel.
c. 1 2 2 :
22y 5 4
y 5 22.
x 2 2 1 2z 5 22
x 1 2z 5 0
x 5 22z.
x 5 22s, y 5 22, z 5 s, sPR.
d. Let z 5 s; sPR.
From 2 :
x 5 y 1 6.
(y 1 6) 1 y 1 2s 5 4
2y 1 2s 5 22
y 5 2s 2 1.
x 5 2s 1 5, y 5 2s 2 1, z 5 s, sPR.
e. 22 ? 2 : 2x 2 4y 2 2z 5 22
Adding 1 :
4x 2 5y 5 0.
5
x 5 y.
4
Let y 5 s, sPR.
5
2a sb 2 s 1 2z 5 2
4
3
s 1 2z 5 2
2
3
z 5 1 2 s.
4
3
5
x 5 s, y 5 s, z 5 1 2 s, sPR
4
4
f. x 2 y 1 2(4) 5 0
x 5 y 2 8.
x 5 s 2 8, y 5 s, z 5 4, sPR.
8. a. The system will have an infinite number of
solutions for any value of k. When k 5 2 equation
2 will be twice that of 1 so the solution is a plane:
x 5 1 2 s 2 2t, y 5 s, z 5 t, s, tPR.
For any other value of k the solution will be a line.
For example k 5 0:
2y 5 24z
y 5 22z.
Chapter 9: Relationships Between Points, Lines, and Planes
x 1 (22z) 1 2z 5 1
x 5 1.
x 5 1, y 5 22s, z 5 s, sPR.
b. No there is no value of k for which the system will
not have a solution. The only time when there is no
solution is when the corresponding coefficients for
each variable differ by a common multiple between
equations, and the constant terms differ by a different
multiple. The only way the first condition is satisfied is
when k 5 2, but when this happens the constant terms
differ by the same factor as the variables, namely 2.
9. The line of intersection of the two planes:
p1: 2x 2 y 1 z 5 0, p2: y 1 4z 5 0 is:
y 5 24z
2x 2 (24z) 1 z 5 0
2x 5 25z
5
x 5 2 z.
2
5
x 5 2 s, y 5 24s, z 5 s, sPR.
2
The direction vector is ( 2 52, 24, 1) or (25, 28, 2).
>
r1 5 s(25, 28, 2), sPR. Since the line we are
looking for is parallel to this line, we know that the
direction vector must be the same. The line passes
through (22, 3, 6) and has direction vector
(25, 28, 2). The equation of the line is
>
r2 5 (22, 3, 6) 1 s(25, 28, 2), sPR.
10. The line of intersection of the two planes,
2x 2 y 1 2z 5 0 and 2x 1 y 1 6z 5 4 is:
4x 1 8z 5 4
x 5 1 2 2z.
2(1 2 2z) 2 y 1 2z 5 0
2 2 y 2 2z 5 0
y 5 2 2 2z.
x 5 1 2 2s, y 5 2 2 2s, z 5 s, sPR.
In order for the a line to be contained in the plane
we need to check that the values for x, y, and z
always satisfy the plane equation:
5x 1 3y 1 16z 2 11 5 0.
5(1 2 2s) 1 3(2 2 2s) 1 16(s) 2 11 5 0
5 1 6 2 11 2 10s 2 6s 1 16s 5 0
0 5 0. Since this is true the line is contained in the
plane.
11. a. p1: 2x 1 y 2 3z 5 3, p2: x 2 2y 1 z 5 21.
p1 2 2p2: 5y 2 5z 5 5
y 5 1 1 z.
2x 1 (1 1 z) 2 3z 5 3
2x 2 2z 5 2
x 5 1 1 z.
x 5 1 1 s, y 5 1 1 s, z 5 s, sPR.
Calculus and Vectors Solutions Manual
b. L meets the xy-plane when z 5 0.
x 5 1, y 5 1. A 5 (1, 1, 0).
L meets the z-axis when both x and y are zero:
s 5 21.
z 5 21.
B 5 (0, 0, 21)
The length of AB is therefore:
"12 1 12 1 12 5 "3 or about 1.73.
12. The line with equation x 5 22y 5 3z has
parametric equations: x 5 s, y 5 2 12 s, z 5 13 s, sPR.
This has the equivalent vector form:
1 1
>
r 5 sa1, 2 , b, sPR.
2 3
The line of intersection of the two planes
x 2 y 1 z 5 1 and 2y 2 z 5 0 is:
1
y5 z
2
1
x2 z1z51
2
1
x 5 1 2 z.
2
1
1
x 5 1 2 2 t, y 5 2 t, z 5 t, tPR. Which has a vector
equation
of:
>
r 5 (1, 0, 0) 1 t (2 12, 12, 1), tPR. The vector
equation of the plane with the given properties is
thus:
1 1
1 1
>
r 5 (1, 0, 0) 1 t a2 , , 1b 1 s a1, 2 , b, s, tPR.
2 2
2 3
The normal vector for the plane is then:
1 1
1 1
1
1 1
a2 , , 1b 3 a1, 2 , b 5 a ? b 2 a1 ? 2 b,
2 2
2 3
2 3
2
1 1
1 1
1
2 7 1
1 ? 1 2 a2 ? b, 2 a2 b 2 ? 1 5 a , , 2 b.
2 3
2 2
2
3 6 4
Or equivalently (8, 14, 23).
The Cartesian equation is then:
8x 1 14y 2 3z 1 D 5 0, and must contain the
point (1, 0, 0).
8(1) 1 D 5 0.
D 5 28.
8x 1 14y 2 3z 2 8 5 0.
Mid-Chapter Review, pp. 518–519
>
1. a. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 15 1 5t
t 5 23
9-11
x 1 (22z) 1 2z 5 1
x 5 1.
x 5 1, y 5 22s, z 5 s, sPR.
b. No there is no value of k for which the system will
not have a solution. The only time when there is no
solution is when the corresponding coefficients for
each variable differ by a common multiple between
equations, and the constant terms differ by a different
multiple. The only way the first condition is satisfied is
when k 5 2, but when this happens the constant terms
differ by the same factor as the variables, namely 2.
9. The line of intersection of the two planes:
p1: 2x 2 y 1 z 5 0, p2: y 1 4z 5 0 is:
y 5 24z
2x 2 (24z) 1 z 5 0
2x 5 25z
5
x 5 2 z.
2
5
x 5 2 s, y 5 24s, z 5 s, sPR.
2
The direction vector is ( 2 52, 24, 1) or (25, 28, 2).
>
r1 5 s(25, 28, 2), sPR. Since the line we are
looking for is parallel to this line, we know that the
direction vector must be the same. The line passes
through (22, 3, 6) and has direction vector
(25, 28, 2). The equation of the line is
>
r2 5 (22, 3, 6) 1 s(25, 28, 2), sPR.
10. The line of intersection of the two planes,
2x 2 y 1 2z 5 0 and 2x 1 y 1 6z 5 4 is:
4x 1 8z 5 4
x 5 1 2 2z.
2(1 2 2z) 2 y 1 2z 5 0
2 2 y 2 2z 5 0
y 5 2 2 2z.
x 5 1 2 2s, y 5 2 2 2s, z 5 s, sPR.
In order for the a line to be contained in the plane
we need to check that the values for x, y, and z
always satisfy the plane equation:
5x 1 3y 1 16z 2 11 5 0.
5(1 2 2s) 1 3(2 2 2s) 1 16(s) 2 11 5 0
5 1 6 2 11 2 10s 2 6s 1 16s 5 0
0 5 0. Since this is true the line is contained in the
plane.
11. a. p1: 2x 1 y 2 3z 5 3, p2: x 2 2y 1 z 5 21.
p1 2 2p2: 5y 2 5z 5 5
y 5 1 1 z.
2x 1 (1 1 z) 2 3z 5 3
2x 2 2z 5 2
x 5 1 1 z.
x 5 1 1 s, y 5 1 1 s, z 5 s, sPR.
Calculus and Vectors Solutions Manual
b. L meets the xy-plane when z 5 0.
x 5 1, y 5 1. A 5 (1, 1, 0).
L meets the z-axis when both x and y are zero:
s 5 21.
z 5 21.
B 5 (0, 0, 21)
The length of AB is therefore:
"12 1 12 1 12 5 "3 or about 1.73.
12. The line with equation x 5 22y 5 3z has
parametric equations: x 5 s, y 5 2 12 s, z 5 13 s, sPR.
This has the equivalent vector form:
1 1
>
r 5 sa1, 2 , b, sPR.
2 3
The line of intersection of the two planes
x 2 y 1 z 5 1 and 2y 2 z 5 0 is:
1
y5 z
2
1
x2 z1z51
2
1
x 5 1 2 z.
2
1
1
x 5 1 2 2 t, y 5 2 t, z 5 t, tPR. Which has a vector
equation
of:
>
r 5 (1, 0, 0) 1 t (2 12, 12, 1), tPR. The vector
equation of the plane with the given properties is
thus:
1 1
1 1
>
r 5 (1, 0, 0) 1 t a2 , , 1b 1 s a1, 2 , b, s, tPR.
2 2
2 3
The normal vector for the plane is then:
1 1
1 1
1
1 1
a2 , , 1b 3 a1, 2 , b 5 a ? b 2 a1 ? 2 b,
2 2
2 3
2 3
2
1 1
1 1
1
2 7 1
1 ? 1 2 a2 ? b, 2 a2 b 2 ? 1 5 a , , 2 b.
2 3
2 2
2
3 6 4
Or equivalently (8, 14, 23).
The Cartesian equation is then:
8x 1 14y 2 3z 1 D 5 0, and must contain the
point (1, 0, 0).
8(1) 1 D 5 0.
D 5 28.
8x 1 14y 2 3z 2 8 5 0.
Mid-Chapter Review, pp. 518–519
>
1. a. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 15 1 5t
t 5 23
9-11
x 5 4 1 2(23), y 5 23 2 3(23),
z 5 15 1 5(23)
x 5 22, y 5 6, z 5 0
(22, 6, 0)
>
b. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 23 2 3t
t 5 21
x 5 4 1 2(21), y 5 23 2 3(21),
z 5 15 1 5(21)
x 5 2, y 5 0, z 5 10
(2, 0, 10)
>
c. r 5 (4, 23, 15) 1 t(2, 23, 5), tPR
x 5 4 1 2t, y 5 23 2 3t, z 5 15 1 5t
0 5 4 1 2t
t 5 22
x 5 4 1 2(22), y 5 23 2 3(22),
z 5 15 1 5(22)
x 5 0, y 5 3, z 5 5
(0, 3, 5)
2. a.–e. Answers may vary. For example:
A(2, 1, 3), B(3, 22, 5), C(28, 25, 7)
a 5 (22.5, 23.5, 6)
b 5 (23, 22, 5)
c 5 (2.5, 20.5, 4)
m1 5 (Aa) 5 (24.5, 24.5, 3) 5 (3, 3, 22)
m2 5 (Bb) 5 (26, 0, 0) 5 (1, 0, 0)
m3 5 (Cc) 5 (10.5, 4.5, 23) 5 (7, 3, 22)
Then substitute in the point and the direction vector
to find the equation of the line.
A(2, 1, 3), B(3, 22, 5), C(28, 25, 7)
m1 5 (Aa) 5 (24.5, 24.5, 3) 5 (3, 3, 22)
m2 5 (Bb) 5 (26, 0, 0) 5 (1, 0, 0)
m3 5 (Cc) 5 (10.5, 4.5, 23) 5 (7, 3, 22)
>
A: r 5 (2, 1, 3) 1 t(3, 3, 22), tPR
x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
>
B: r 5 (3, 22, 5) 1 t(1, 0, 0), tPR
x 5 3 1 t, y 5 22, z 5 5, tPR
>
C: r 5 (28, 25, 7) 1 t(7, 3, 22), tPR
x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
A: x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
B: x 5 3 1t, y 5 22, z 5 5, tPR
C: x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
y 5 22 5 1 1 3t
t 5 21
x 5 2 1 3(21), y 5 1 1 3(21),
z 5 3 2 2(21)
x 5 21, y 5 22, z 5 5
(21, 22, 5)
9-12
A: x 5 2 1 3t, y 5 1 1 3t, z 5 3 2 2t, tPR
B: x 5 3 1 t, y 5 22, z 5 5, tPR
C: x 5 28 1 7t, y 5 25 1 3t, z 5 7 2 2t, tPR
y 5 22 5 25 1 3t
t51
x 5 28 1 7(1), y 5 25 1 3(1), z 5 7 2 2(1)
x 5 21, y 5 22, z 5 5
(21, 22, 5)
The three medians meet at (21, 22, 5).
3. a. L1: 5x 1 y 1 2z 1 15 5 0
L2: 4x 1 y 1 2z 1 8 5 0
L1 2 L2: x 1 7 5 0
So x 5 27.
L1: y 1 2z 2 20 5 0
L2: y 1 2z 2 20 5 0
z 5 t,
y 1 2(t) 2 20 5 0
y 5 20 2 2t
>
r 5 (27, 20, 0) 1 t(0, 22, 1), tPR
b.
L1: 4x 1 3y 1 3z 2 2 5 0
L2: 5x 1 2y 1 3z 1 5 5 0
2L1 2 3L2: 27x 2 3z 2 19 5 0
z 5 7t,
27x 2 3(7t) 2 19 5 0,
19
x 5 23t 2
7
19
4a23t 2 b 1 3y 1 3(7t) 2 2 5 0
7
30
y 5 23t 1
7
19 30
>
r 5 a2 , , 0b 1 t(3, 3, 27), tPR
7 7
>
c. L1: r 5 (27, 20, 0) 1 t(0, 22, 1), tPR
19 30
>
L2: r 5 a2 , , 0b 1 t(3, 3, 27), tPR
7 7
L1: x 5 27, y 5 20 2 2t, z 5 t
19
30
L2: x 5 2 1 3t, y 5
1 3t, z 5 27t
7
7
30
19
2 1 3t 5 27, t 5 2
7
21
19
30
30
30
x 5 2 1 3a2 b, y 5
1 3a2 b,
7
21
7
21
30
z 5 27a2 b
21
x 5 27, y 5 0, z 5 10
(27, 0, 10)
4. a. p1: 3x 1 y 1 7z 1 3 5 0
p2: x 2 13y 2 3z 2 38 5 0
Chapter 9: Relationships Between Points, Lines, and Planes
13p1 1 p2: 40x 1 88z 1 1 5 0
z 5 t,
40x 1 88(t) 1 1 5 0
11t
1
x52
2
5
40
11t
1
3a2
2 b 1 y 1 7(t) 1 3 5 0
5
40
2t
117
y52 2
5
40
2t
11t
1
117
x52
2 ,y52 2
, z 5 t, tPR
5
40
5
40
p1: x 2 3y 1 z 1 11 5 0
b.
p2: 6x 2 13y 1 8z 2 28 5 0
26p1 1 p2: 5y 1 2z 2 94 5 0
z 5 s,
5y 1 2(s) 2 94 5 0
2
94
y52 s1
5
5
2
94
x 2 3a2 s 1 b 1 (s) 1 11 5 0
5
5
227
11
x52 s1
5
5
2
11
227
94
x52 s1
, y 5 2 s 1 , z 5 s, sPR
5
5
5
5
c. The lines found in 4. a. and 4. b. do not intersect,
because they are in parallel planes.
5. a. For there to be no solution the lines must be
inconsistent with each other.
L1: x 1 ay 5 9
L2: ax 1 9y 5 227
1
a
5
a
9
a 5 63
For a 5 3:
L1: x 1 3y 5 9
L2: 3x 1 9y 5 227
For a 5 23, the equations are equivalent.
So there is no solution when a 5 3.
b. To have an infinite number of solutions, the lines
must be proportional.
L1: x 1 ay 5 9
L2: ax 1 9y 5 227
23(x 1 ay 5 9) 5 23x 2 3ay 5 227
L1: 23x 2 3ay 5 227
L2: ax 1 9y 5 227
a 5 23
c. The system has one solution when a 2 3 or
a 2 23, because other values lead to an infinite
number of solutions or no solution.
Calculus and Vectors Solutions Manual
x 2 11
y24
z 2 27
5
5
5s
2
24
5
L2: x 5 0, y 5 1 2 3t, z 5 3 1 2t, tPR
L1: x 5 2s 1 11, y 5 24s 1 4, z 5 27 1 5s
x 5 0 5 2s 1 11,
s 5 25.5
y 5 24(25.5) 1 4, z 5 27 1 5(25.5)
x 5 0, y 5 26, z 5 20.5
25
y 5 26 5 1 2 3t, t 5 2
3
7
z 5 20.5 5 3 1 2t, t 5 2
4
Since there is no t-value that satisfies the equations,
there is no intersection, and these lines are skew.
x25
z14
7. a. L1:
5y225
5s
2
23
L2: (x 2 3, y 2 20, z 2 7) 5 t(2, 24, 5), tPR
L1: x 5 2s 1 5, y 5 s 1 2, z 5 23s 2 4
L2: x 5 2t 1 3, y 5 24t 1 20, z 5 5t 1 7
x 5 2t 1 3 5 2s 1 5
y 5 s 1 2 5 24t 1 20
z 5 23s 2 4 5 5t 1 7
L3: 2t 2 2s 2 2 5 0
L4: 4t 1 s 2 18 5 0
L5: 5t 1 3s 1 11 5 0
L3 1 2L4: 10t 2 38 5 0, t 5 3.8
3L3 1 2L5: 16t 1 16 5 0, t 5 21
b. Since there is no t-value that satisfies the
equations, there is no intersection, and these lines
are skew.
8. L1: x 5 1 1 2s, y 5 4 2 s, z 5 23s, sPR
L2: x 5 23, y 5 t 1 3, z 5 2t, tPR
x 5 23 5 1 1 2s
s 5 22
x 5 23, y 5 6, z 5 6
(23, 6, 6)
>
9. a. L1: r 5 (5, 1, 7) 1 s(2, 0, 5), sPR
>
L2: r 5 (21, 21, 3) 1 t(4, 2, 21), tPR
L1: x 5 5 1 2s, y 5 1, z 5 7 1 5s
L2: x 5 21 1 4t, y 5 21 1 2t, z 5 3 2 t
y 5 1 5 21 1 2t,
t51
x 5 21 1 4(1), y 5 21 1 2(1),
z 5 3 2 (1)
x 5 3, y 5 1, z 5 2
(3, 1, 2)
>
b. L1: r 5 (2, 21, 3) 1 s(5, 21, 6), sPR
>
L2: r 5 (28, 1, 29) 1 t(5, 21, 6), tPR
These lines are the same, so either one of these
lines can be used as their intersection.
6. L1:
9-13
10. a. Answers may vary. For example:
i. coincident
z
L1
L2
y
x
ii. parallel and distinct
z
L2
L1
y
x
iii. skew
z
L1
x
y
L2
iv. intersect in a point
z
L1
L2
y
x
b. i. When lines are the same, they are a multiple
of each other.
ii. When lines are parallel, one equation is a
multiple of the other equation, except for the
constant term.
9-14
iii. When lines are skew, there are no common
solutions to make each equation consistent.
iv. When the solution meets in a point, there is only
one unique solution for the system.
11. a. A line and plane have an infinite number of
points of intersection when the line lies in the plane.
b. Answers may vary. For example:
>
r 5 t(3, 25, 23), tPR
>
r 5 t(3, 25, 23) 1 s(1, 1, 1), t, sPR
12. a. 1 2x 1 3y 5 30
2
x 2 2y 5 213
Equation 1 2 (2 3 equation 2 ): 7y 5 56
y58
2x 1 24 5 30
x53
(3, 8)
b. 1
x 1 4y 2 3z 1 6 5 0
2 2x 1 8y 2 6z 1 11 5 0
There is no solution to this system, because the
planes are parallel, but one plane lies above the
other.
c. 1
x 2 3y 2 2z 5 29
2
2x 2 5y 1 z 5 3
3 23x 1 6y 1 2z 5 8
Equation 1 1 (2 3 equation 2 ): 5x 2 13y 5 23
Equation 2 1 ( equation 3 ): 22x 1 3y 5 21
2(5x 2 13y 5 23)
1 5(22x 1 3y 5 21)
211y 5 211
y51
5x 2 13(1) 5 23
x52
(2) 2 3(1) 2 2z 5 29
z54
(2, 1, 4)
13. a. The two lines intersect at a point.
b. The two planes are parallel and do not meet.
c. The three planes intersect at a point.
14. a. L:(x 2 y 5 1) 1 (y 1 z 5 23)
5 x 1 z 5 22
1
L1: y 2 z 5 0, x 5 2
2
x 1 z 5 22
1
a2 b 1 z 5 22
2
3
z52
2
y2z50
Chapter 9: Relationships Between Points, Lines, and Planes
3
y 2 a2 b 5 0
2
y52
3
2
1 3 3
a2 , 2 , 2 b
2 2 2
0 n ? n1 0
b. cos u 5
0 n 0 0 n1 0
n 5 (1, 1, 21)
n1 5 (0, 1, 1)
0
cos u 5
@ "3 @ @ "2 @
u 5 90 °
c. (0, 1, 1) 3 (1, 1, 21) 5 (22, 1, 21)
5 (2, 21, 1)
Ax 1 By 1 Cz 1 D 5 0
2x 2 y 1 z 1 D 5 0
21
23
23
2a
b2a
b1a
b1D50
2
2
2
D51
2x 2 y 1 z 1 1 5 0
9.4 The Intersection of Three Planes,
pp. 531–533
x 2 3y 1 z 5 2
0x 1 y 2 z 5 21
3 0x 1 0y 1 3z 5 212
The system can be solved by first solving equation 3
for z. Thus,
3z 5 212
z 5 24
If we use the method of back substitution, we can
substitute z 5 24 into equation 2 and solve for y.
y 2 (24) 5 21
y 5 25
If we substitute y 5 25 and z 5 24 into equation 1
we obtain the value of x.
x 2 3(25) 2 4 5 2 or x 5 29
The three planes intersect at the point with
coordinates (29, 25, 24)
Check:
Substituting into equation 1 :
x 2 3y 1 z 5 29 1 15 2 4 5 2
Substituting into equation 2 :
0x 1 y 2 z 5 25 1 4 5 21
Substituting into equation 3 : 0x 1 0y 1 3z 5 212
1. a.
1
2
Calculus and Vectors Solutions Manual
b. This solution is the point at which all three
planes meet.
2. a. 1
x2y1z54
2 0x 1 0y 1 0z 5 0
3 0x 1 0y 1 0z 5 0
The answer may vary depending upon the constant
you multiply the equations by. For example,
2 3 (x 2 y 1 z 5 4) 5 2x 2 2y 1 2z 5 8
3 3 (x 2 y 1 z 5 4) 5 3x 2 3y 1 3z 5 12
3x 2 3y 1 3z 5 12 and 2x 2 2y 1 2z 5 8 are
equations that could work.
b. These three planes are intersecting in one single
plane, because all three equations can be changed
into one equivalent equation. They are coincident
planes.
c. Setting x 5 t and y 5 s leads to
t 2 s 1 z 5 4 or z 5 s 2 t 1 4, s, tPR
d. Setting y 5 t and z 5 s leads to
x 2 t 1 s 5 4 or x 5 t 2 s 1 4, s, tPR
3. a. 1 2x 2 y 1 3z 5 22
2
x 2 y 1 4z 5 3
3 0x 1 0y 1 0z 5 1
The answer may vary depending upon the constants
and equations you use to determine your answer.
For example,
Equation 1 1 equation 2 1 equation 3 5
(2x 2 y 1 3z 5 22)
1 (x 2 y 1 4z 5 3)
1 (0x 1 0y 1 0z 5 1)
3x 2 2y 1 7z 5 2
or
2 3 equation 2 2 equation 3 5
(2x 2 2y 1 8z 5 6)
2 (0x 1 0y 1 0z 5 1)
2x 2 2y 1 8z 5 5
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
3x 2 2y 1 7z 5 2 is one system of equations that
could produce the original system composed of
equations 1 , 2 , and 3 .
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
2x 2 2y 1 8z 5 5 is another system of equations
that could produce the original system composed of
equations 1 , 2 , and 3 .
b. The systems have no solutions.
4. a. 1
x 1 2y 2 z 5 4
2 x 1 0y 2 2z 5 0
3 2x 1 0y 1 0z 5 26
The system can be solved by first solving equation 3
for x. So,
9-15
3
y 2 a2 b 5 0
2
y52
3
2
1 3 3
a2 , 2 , 2 b
2 2 2
0 n ? n1 0
b. cos u 5
0 n 0 0 n1 0
n 5 (1, 1, 21)
n1 5 (0, 1, 1)
0
cos u 5
@ "3 @ @ "2 @
u 5 90 °
c. (0, 1, 1) 3 (1, 1, 21) 5 (22, 1, 21)
5 (2, 21, 1)
Ax 1 By 1 Cz 1 D 5 0
2x 2 y 1 z 1 D 5 0
21
23
23
2a
b2a
b1a
b1D50
2
2
2
D51
2x 2 y 1 z 1 1 5 0
9.4 The Intersection of Three Planes,
pp. 531–533
x 2 3y 1 z 5 2
0x 1 y 2 z 5 21
3 0x 1 0y 1 3z 5 212
The system can be solved by first solving equation 3
for z. Thus,
3z 5 212
z 5 24
If we use the method of back substitution, we can
substitute z 5 24 into equation 2 and solve for y.
y 2 (24) 5 21
y 5 25
If we substitute y 5 25 and z 5 24 into equation 1
we obtain the value of x.
x 2 3(25) 2 4 5 2 or x 5 29
The three planes intersect at the point with
coordinates (29, 25, 24)
Check:
Substituting into equation 1 :
x 2 3y 1 z 5 29 1 15 2 4 5 2
Substituting into equation 2 :
0x 1 y 2 z 5 25 1 4 5 21
Substituting into equation 3 : 0x 1 0y 1 3z 5 212
1. a.
1
2
Calculus and Vectors Solutions Manual
b. This solution is the point at which all three
planes meet.
2. a. 1
x2y1z54
2 0x 1 0y 1 0z 5 0
3 0x 1 0y 1 0z 5 0
The answer may vary depending upon the constant
you multiply the equations by. For example,
2 3 (x 2 y 1 z 5 4) 5 2x 2 2y 1 2z 5 8
3 3 (x 2 y 1 z 5 4) 5 3x 2 3y 1 3z 5 12
3x 2 3y 1 3z 5 12 and 2x 2 2y 1 2z 5 8 are
equations that could work.
b. These three planes are intersecting in one single
plane, because all three equations can be changed
into one equivalent equation. They are coincident
planes.
c. Setting x 5 t and y 5 s leads to
t 2 s 1 z 5 4 or z 5 s 2 t 1 4, s, tPR
d. Setting y 5 t and z 5 s leads to
x 2 t 1 s 5 4 or x 5 t 2 s 1 4, s, tPR
3. a. 1 2x 2 y 1 3z 5 22
2
x 2 y 1 4z 5 3
3 0x 1 0y 1 0z 5 1
The answer may vary depending upon the constants
and equations you use to determine your answer.
For example,
Equation 1 1 equation 2 1 equation 3 5
(2x 2 y 1 3z 5 22)
1 (x 2 y 1 4z 5 3)
1 (0x 1 0y 1 0z 5 1)
3x 2 2y 1 7z 5 2
or
2 3 equation 2 2 equation 3 5
(2x 2 2y 1 8z 5 6)
2 (0x 1 0y 1 0z 5 1)
2x 2 2y 1 8z 5 5
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
3x 2 2y 1 7z 5 2 is one system of equations that
could produce the original system composed of
equations 1 , 2 , and 3 .
2x 2 y 1 3z 5 22, x 2 y 1 4z 5 3, and
2x 2 2y 1 8z 5 5 is another system of equations
that could produce the original system composed of
equations 1 , 2 , and 3 .
b. The systems have no solutions.
4. a. 1
x 1 2y 2 z 5 4
2 x 1 0y 2 2z 5 0
3 2x 1 0y 1 0z 5 26
The system can be solved by first solving equation 3
for x. So,
9-15
2x 5 26
x 5 23
If we use the method of back substitution, we can
substitute x 5 23 into equation 2 and solve for z.
23 2 2z 5 0
3
z52
2
If we substitute x 5 23 and z 5 2 32 into equation 1
we obtain the value of y.
11
3
23 1 2y 1 5 4 or y 5
2
4
The equations intersect at the point with coordinates
(23, 114, 2 32 )
Check:
Substituting into equation 1 :
x 1 2y 2 z 5 23 1 224 1 32 5 4
Substituting into equation 2 :
x 1 0y 2 2z 5 23 1 3 5 0
Substituting into equation 3 : 2x 1 0y 1 0z 5 26
b. This solution is the point at which all three
planes meet.
5. a. 1
2x 2 y 1 z 5 1
2
x 1 y 2 z 5 21
3 23x 2 3y 1 3z 5 3
Since equation 3 5 2 equation 2 , equation 2
and equation 3 are consistent or lie in the same
plane. Equation 1 meets this plane in a line.
b. Adding equation 2 and equation 1 creates an
equivalent equation, 3x 5 0 or x 5 0. Substituting
x 5 0 into equation 1 and equation 2 gives
equation 4 z 2 y 5 1 and equation 5
y 2 z 5 21. Equations 4 and 5 indicate the
problem has infinite solutions. Substituting y 5 t
into equation 4 or 5 leads to
x 5 0, y 5 t, and z 5 1 1 t, tPR
Check:
2(0) 2 s 1 (s 1 1) 5 1
0 1 s 2 (s 1 1) 5 21
23(0) 2 3(s) 1 3(s 1 1) 5 3
6. 1 2x 1 3y 2 4z 5 25
2
x 2 y 1 3z 5 2201
3 5x 2 5y 1 15z 5 21004
There is no solution to this system of equations,
because if you multiply equation 2 by 5 you
obtain a new equation, 5x 2 5y 1 15z 5 21005,
which is inconsistent with equation 3 .
7. a. Yes when this equation is alone, this is true,
because any constants can be substituted into the
variables in the equation 0x 1 0y 1 0z 5 0 and the
equation will always be consistent.
9-16
b. Answers may vary. For example: To obtain a no
solution and an equation with 0x 1 0y 1 0z 5 0,
you must have two equal planes and one parallel
distinct plane. For example one solution is:
x1y1z52
2x 1 2y 1 2z 5 4
3x 1 3y 1 3z 5 12
8. a. 1 2x 1 y 2 z 5 23
2 x 2 y 1 2z 5 0
3 3x 1 2y 2 z 5 25
2 3 equation 2 1 equation 3 5 5x 1 0y 1 0z
5 25 which gives x 5 21.
Equation 1 1 equation 2 5 3x 1 0y 1 1z
5 23. Substituting x 5 1 into this equation leads
to: 3(21) 1 z 5 23 or z 5 0.
Substituting z 5 0 and x 5 21 into equation 1
gives: 2(21)y 2 0 5 23 or y 5 21. (21, 21, 0)
is the point at which the three planes meet.
Check:
Substituting into equation 1 :
2x 1 y 2 z 5 22 2 1 1 0 5 23
Substituting into equation 2 :
x 2 y 1 2z 5 21 1 1 1 0 5 0
Substituting into equation 3 :
3x 1 2y 2 z 5 23 2 2 1 0 5 25
x
y
7
b. 1
2 1z5
3
4
8
2 2x 1 2y 2 3z 5 220
3
x 2 2y 1 3z 5 2
Equation 2 1 equation 3 5 3x 1 0y 1 0z 5 218
which gives x 5 26.
Equation 3 2 3 3 Equation 1 5 2 54y 5 2 58 or
y 5 12. Substituting x 5 26 and y 5 12 into equation 3
leads to:
1
26 2 2a b 1 3z 5 2 or z 5 3.
2
(26, 12, 3) is the point at which the three planes meet.
Check:
Substituting into equation 1 :
x
y
1
7
2 4 1 z 5 22 2 8 1 3 5 8
3
Substituting into equation 2 :
2x 1 2y 2 3z 5 212 1 1 2 9 5 220
Substituting into equation 3 :
x 2 2y 1 3z 5 26 2 1 1 9 5 2
c. 1 x 2 y 5 2199
2 x 1 z 5 2200
3 y 2 z 5 201
Equation 2 1 equation 3 5 equation 4
5x1y51
Chapter 9: Relationships Between Points, Lines, and Planes
Equation 4 1 equation 1 5 2x 5 2198 or
x 5 299. Substituting x 5 299 into equation 1
leads to:
299 2 y 5 2199 or y 5 100. Substituting x 5 299
into equation 2 , you obtain:
299 1 z 5 2200 or z 5 2101
(299, 100, 2101) is the point at which the three
planes meet.
Check:
Substituting into equation 1 :
x 2 y 5 299 2 (100) 5 2199
Substituting into equation 2 :
x 1 z 5 299 2 101 5 2200
Substituting into equation 3 :
y 2 z 5 100 2 (2101) 5 201
d. 1 x 2 y 2 z 5 21
2
y2250
3
x1155
Rearranging equation 2 gives y 5 2. Solving for x
in equation 3 gives x 5 4.
Substituting x 5 4 and y 5 2 into equation 1
leads to:
4 2 2 2 z 5 21 or z 5 3.
(4, 2, 3) is the point at which all three planes meet.
9. a. 1 x 2 2y 1 z 5 3
2 2x 1 3y 2 z 5 29
3 5x 2 3y 1 2z 5 0
Equation 3 1 equation 2 5 equation 4
5 7x 1 1z 5 29.
Setting z 5 t, x 5 2 17 t 2 97
Equation 2 2 2 3 equation 1 5 equation 5
5 7y 1 23z 5 215.
Setting z 5 t, y 5 2 157 1 37 t
x 5 2 17 t 2 97, y 5 2 157 1 37 t, and z 5 t, tPR The
planes intersect in a line.
b. 1 x 2 2y 1 z 5 3
2
x1y1z52
3 x 2 3y 1 z 5 26
Equation 3 2 equation 2 5 24y 5 28 or y 5 2
Equation 3 2 equation 1 5 21y 5 29 or y 5 9
Since the solutions for y are different from these
two equations, there is no solution to this system of
equations.
c. 1 x 2 y 1 z 5 22
2 x1y1z52
3 x 2 3y 1 z 5 26
Equation 1 1 equation 2 5 equation 4
5 2x 1 2y 5 0.
Setting z 5 t, x 5 2t
Calculus and Vectors Solutions Manual
Using z 5 t and x 5 2t, Solve equation 1
2t 2 y 1 t 5 22 or y 5 2
x 5 2t, y 5 2, and z 5 t, tPR
The planes intersect in a line.
10. a. 1
x2y1z52
2 2x 2 2y 1 2z 5 4
3
x 1 y 2 z 5 22
Equation 1 1 equation 3 5 equation 4
5 2x 5 0 or x 5 0.
Setting z 5 t,
Equation 1 : 0 2 y 1 t 5 2 or y 5 t 2 2
x 5 0, y 5 t 2 2, and z 5 t, tPR
b. 1 2x 2 y 1 3z 5 0
2 4x 2 2y 1 6z 5 0
3 22x 1 y 2 3z 5 0
Equation 1 1 equation 3 5 equation 4
5 2x 5 0 or x 5 0.
Setting y 5 t and z 5 s, equation 1 :
t 2 3s
2x 2 t 1 3s 5 0 or x 5
2
t 2 3s
x5
, y 5 t, and z 5 s, s, tPR
2
x1y1z51
x 2 2y 1 z 5 0
3
x2y1z50
1
Equation
2 equation 3 5 equation 4
5 2y 5 1 or y 5 12
Equation 2 2 equation 3 5 equation 5
5 2y 5 0 or y 5 0
Since the y-variable is different in equation 4 and
equation 5 , the system is inconsistent and has no
solution.
b. Answers may vary. For example: If you use the
normals from equations 1 , 2 , and 3 , you can
determine the direction vectors from the equations’
coefficients.
>
n1 5 (1, 1, 1)
>
n2 5 (1, 22, 1)
>
n3 5 (1, 21, 1)
>
>
m1 5 n1 3 n2 5 (3, 0, 23)
>
>
m2 5 n1 3 n3 5 (2, 0, 22)
>
>
m3 5 n2 3 n3 5 (21, 0, 1)
c. The three lines of intersection are parallel and are
pairwise coplanar, so they form a triangular prism.
>
>
>
d. n1 3 n2 is perpendicular to n3 . So since,
>
>
>
(n1 3 n2 ) ? n3 5 0, a triangular prism forms.
12. a. 1 x 2 y 1 3z 5 3
2 x 2 y 1 3z 5 6
3
3x 2 5z 5 0
11. a.
1
2
9-17
Equation 1 and equation 2 have the same set of
coefficients and variables, however, equations 1
equals 3 while equation 2 equals 6, which means
there is no possible solution.
b. 1 5x 2 2y 1 3z 5 1
2 5x 2 2y 1 3z 5 21
3 5x 2 2y 1 3z 5 13
All three equations equal different numbers so there
is no possible solution.
c. 1
x2y1z59
2 2x 2 2y 1 2z 5 18
3 2x 2 2y 1 2z 5 17
Equation 2 equals 18 while equation 3 equals 17,
which means there is no possible solution.
d. The coefficients of equation 1 are half the
coefficients of equation 2 , but the constant term
is not half the other constant term.
13. a. 1 2x 2 y 2 z 5 10
2 x 1 y 1 0z 5 7
3 0x 1 y 2 z 5 8
Equation 1 2 2 3 equation 2 2 equation 3 :
24y 5 2 12 or y 5 3. Substituting y 5 3 into
equation 2 and equation 3 gives:
x 1 3 1 0z 5 7 or x 5 4
0x 1 3 2 z 5 8 or z 5 25
(4, 3, 25)
b. 1 2x 2 y 1 z 5 23
2
x 1 y 2 2z 5 1
3 5x 1 2y 2 5z 5 0
Equation 1 1 equation 2 : 3x 2 z 5 22.
Setting z 5 t, x 5
Equation
Setting
1
t22
3
2 2 3 equation
2
: 23y 1 5z 5 25.
5t 1 5
z 5 t, y 5 3
t22
5t 1 5
,y5
, z 5 t, tPR
3
3
c. 1
x1y2z50
2
2x 2 y 1 z 5 0
3 4x 2 5y 1 5z 5 0
Equation 1 1 equation 2 : 3x 5 0 or x 5 0
Setting x 5 0 and z 5 t in equation 2 gives,
2(0) 2 y 1 t 5 0 or y 5 t
x 5 0, y 5 t, z 5 t, tPR
d. 1 x 2 10y 1 13z 5 24
2 2x 2 20y 1 26z 5 28
3 x 2 10y 1 13z 5 28
If you multiply equation 2 by two, you obtain
2x 2 20y 1 26z 5 216. Since equation 2 and
x5
9-18
equation 3 equal different numbers, there is no
solution to this system.
e. 1 x 2 y 1 z 5 22
2
x1y1z52
3 3x 1 y 1 3z 5 2
Equation 1 1 equation 2 : 22y 5 24 or y 5 2
Setting y 5 2 and z 5 t in equation 1 ,
x 2 2 1 t 5 22 or x 5 2t
x 5 2t, y 5 2, z 5 t, tPR
f. 1 x 1 y 1 z 5 0
2 x 2 2y 1 3z 5 0
3 2x 2 y 1 3z 5 0
Equation 1 2 equation 2 5 equation 4
5 3y 2 2z 5 0
Equation 3 2 2 3 equation 2 2 equation 5
5 3y 2 3z 5 0
Equation 4 2 equation 5 : z 5 0
Setting z 5 0 in equation 1 and equation 2 ,
Equation 6 5 x 1 y 5 0
Equation 7 5 x 2 2y 5 0
Equation 6 2 equation 7 : 3y 5 0 or y 5 0
Setting y 5 0 and z 5 0 in equation 1 leads to
x50
(0, 0, 0)
14. a. First, reorder these equations so that equation
2 is first, equation 3 is second, and equation 1
last.
1
x2y1z5p
2 4x 1 qy 1 z 5 2
3 2x 1 y 1 z 5 4
To eliminate x from the last two equations, subtract
4 times equation 1 from equation 2 , and subtract
2 times equation 1 from equation 3 .
1
x2y1z5p
2 (q 1 4)y 2 3z 5 2 2 4p
3
3y 2 z 5 4 2 2p
There will be an infinite number of solutions if
q 1 4 5 9 and 3(4 2 2p) 5 2 2 4p because then
equation 2 will be 3 times equation 3 . This means
that p 5 q 5 5.
b. Based on what was found in part a., substituting
in p 5 q 5 5 we will arrive at the equivalent system
1 x2y1z55
2
9y 2 3z 5 218
3
3y 2 z 5 26
which is really the same as
1 x2y1z55
2
3y 2 z 5 26
Letting z 5 t, we see that equation 2 delivers
Chapter 9: Relationships Between Points, Lines, and Planes
1
(t 2 6)
3
1
5 t22
3
and so equation 1 gives
1
x 5 (t 2 6) 2 t 1 5
3
2
52 t13
3
So the parametric equation of the line of
intersection is
2
1
x 5 2 t 1 3, y 5 t 2 2, z 5 t, tPR.
3
3
15. a. First, eliminate x from two of these equations.
To make things easier, switch equation 1 with
equation 2 , and multiply equation 3 by 2.
1
2x 1 y 1 z 5 24
2
4x 1 3y 1 3z 5 28
3 6x 2 4y 1 (2m 2 2 12)z 5 2m 2 8
Now eliminate x from the last two equations by
using proper multiples of the first equation.
1
2x 1 y 1 z 5 24
2
y1z50
3 27y 1 (2m 2 2 15)z 5 2m 1 4
Now eliminate y from the third equation by using a
proper multiple of the second equation.
1 2x 1 y 1 z 5 24
2
y1z50
3 (2m 2 2 8)z 5 2m 1 4
If 2m 2 2 8 5 0 (the coefficient of z in the third
equation), then m 5 6 2. However, if m 5 2, the
third equation would become 0z 5 8, which has no
solutions. So there is no solution if m 5 2.
b. Working with what was found in part a., if m 2 62,
then the third equation in the equivalent system found
there will have a unique solution for z, namely
2m 1 4
z5
,
2m 2 2 8
and back-substituting into the other two equations
will give unique solutions for x and y also. So there
is a unique solution if m 2 62.
c. Again using the equivalent system found in part a.,
setting m 5 22 will deliver the third equation 0z 5 0,
which allows for z to be anything at all. So m 5 22
will give an infinite number of solutions.
1
1
1
16. a. 1
1 2 50
a
c
b
2
2
3
13
2
1 1 5
a
c
b
6
y5
Calculus and Vectors Solutions Manual
4
3
2
5
2 1 5
a
c
b
2
Equation 2 2 2 3 equation
1
4
13
1 5
5 equation 4
c
b
6
Equation 3 2 4 3 equation
3
1
:
1
6
7
:2 1
c
b
5
>
>
m3 5 n 3 n1 5 (21, 0, 1) 5 5 equation
2
Equation 5 1 6 3 equation 4 :
31
5 15.5 or c 5 2
c
Substituting c 5 2 into equation 4 :
13
1
or b 5 6
125
b
6
Substituting c 5 2 and b 5 6 into equation
1
1
1
1 2 5 0 or a 5 3
a
6
2
(3, 6, 2)
5
1
:
9.5 The Distance from a Point to a
Line in R2 and R3, pp. 540–541
1. a. 3x 1 4y 2 5 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(24) 1 4(5) 2 5 0
d5
"32 1 42
3
5
5
b. 5x 2 12y 1 24 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 5(24) 2 12(5) 1 24 0
d5
"52 1 (212)2
56
or 4.31
5
13
c. 9x 2 40y 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 9(24) 2 40(5) 0
d5
"92 1 (40)2
236
5
or 5.76
"1681
2. a. 2x 2 y 1 1 5 0 and 2x 2 y 1 6 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
9-19
1
(t 2 6)
3
1
5 t22
3
and so equation 1 gives
1
x 5 (t 2 6) 2 t 1 5
3
2
52 t13
3
So the parametric equation of the line of
intersection is
2
1
x 5 2 t 1 3, y 5 t 2 2, z 5 t, tPR.
3
3
15. a. First, eliminate x from two of these equations.
To make things easier, switch equation 1 with
equation 2 , and multiply equation 3 by 2.
1
2x 1 y 1 z 5 24
2
4x 1 3y 1 3z 5 28
3 6x 2 4y 1 (2m 2 2 12)z 5 2m 2 8
Now eliminate x from the last two equations by
using proper multiples of the first equation.
1
2x 1 y 1 z 5 24
2
y1z50
3 27y 1 (2m 2 2 15)z 5 2m 1 4
Now eliminate y from the third equation by using a
proper multiple of the second equation.
1 2x 1 y 1 z 5 24
2
y1z50
3 (2m 2 2 8)z 5 2m 1 4
If 2m 2 2 8 5 0 (the coefficient of z in the third
equation), then m 5 6 2. However, if m 5 2, the
third equation would become 0z 5 8, which has no
solutions. So there is no solution if m 5 2.
b. Working with what was found in part a., if m 2 62,
then the third equation in the equivalent system found
there will have a unique solution for z, namely
2m 1 4
z5
,
2m 2 2 8
and back-substituting into the other two equations
will give unique solutions for x and y also. So there
is a unique solution if m 2 62.
c. Again using the equivalent system found in part a.,
setting m 5 22 will deliver the third equation 0z 5 0,
which allows for z to be anything at all. So m 5 22
will give an infinite number of solutions.
1
1
1
16. a. 1
1 2 50
a
c
b
2
2
3
13
2
1 1 5
a
c
b
6
y5
Calculus and Vectors Solutions Manual
4
3
2
5
2 1 5
a
c
b
2
Equation 2 2 2 3 equation
1
4
13
1 5
5 equation 4
c
b
6
Equation 3 2 4 3 equation
3
1
:
1
6
7
:2 1
c
b
5
>
>
m3 5 n 3 n1 5 (21, 0, 1) 5 5 equation
2
Equation 5 1 6 3 equation 4 :
31
5 15.5 or c 5 2
c
Substituting c 5 2 into equation 4 :
13
1
or b 5 6
125
b
6
Substituting c 5 2 and b 5 6 into equation
1
1
1
1 2 5 0 or a 5 3
a
6
2
(3, 6, 2)
5
1
:
9.5 The Distance from a Point to a
Line in R2 and R3, pp. 540–541
1. a. 3x 1 4y 2 5 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(24) 1 4(5) 2 5 0
d5
"32 1 42
3
5
5
b. 5x 2 12y 1 24 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 5(24) 2 12(5) 1 24 0
d5
"52 1 (212)2
56
or 4.31
5
13
c. 9x 2 40y 5 0
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 9(24) 2 40(5) 0
d5
"92 1 (40)2
236
5
or 5.76
"1681
2. a. 2x 2 y 1 1 5 0 and 2x 2 y 1 6 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
9-19
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
2(0) 2 y 1 1 5 0 or y 5 1 which corresponds to
the point (0, 1).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 2(0) 2 1(1) 1 6 0
d5
"22 1 (21)2
5
or 2.24
5
"5
b. 7x 2 24y 1 168 5 0 and 7x 2 24y 2 336 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
7(0) 2 24y 1 168 5 0 or y 5 7 which corresponds
to the point (0, 7)
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 7(0) 2 24(7) 2 336 0
d5
"72 1 (224)2
504
or 20.16
5
25
>
3. a. r 5 (21, 2) 1 s(3, 4), sPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 21 1 3s,
y 5 2 1 4s. We construct a vector from R(22, 3)
to a general point on the line.
>
a 5 322 2 (21 1 3s), 3 2 (2 1 4s)4
5 (21 2 3s, 1 2 4s).
(3, 4) ? (21 2 3s, 1 2 4s) 5 0
(23 2 9s) 1 (4 2 16s) 5 0
1
s5
25
This means that the minimal distance between
R(22, 3) and the line occurs when s 5 251 .
54
This point corresponds to A2 22
25 , 25 B. The distance
between this point and (22, 3) is 1.4.
>
b. r 5 (1, 0) 1 t(5, 12), tPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 5t,
y 5 12t. We construct a vector from R(22, 3) to a
general point on the line.
>
a 5 322 2 (1 1 5t), 3 2 (12t)4
5 (23 2 5t, 3 2 12t).
(5, 12) ? (23 2 5t, 3 2 12t) 5 0
9-20
(215 2 25t) 1 (36 2 144t) 5 0
21
t5
169
This means that the minimal distance between
21
R(22, 3) and the line occurs when t 5 169
.
274 252
This point corresponds to A 169, 169 B. The distance
between this point and (22, 3) is about 3.92.
>
c. r 5 (1, 3) 1 p(7, 224), pPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 7p,
y 5 3 2 24p. We construct a vector from R(22, 3)
to a general point on the line.
>
a 5 322 2 (1 1 7p), 3 2 (3 2 24p)4
5 (23 2 7p, 24p).
(7, 224) ? (23 2 7p, 24p) 5 0
(221 2 49p) 1 (2576p) 5 0
21
p52
625
This means that the minimal distance between
21
R(22, 3) and the line occurs when p 5 2 625
.
478 2379
This point corresponds to A 625, 625 B.
The distance between this point and (22, 3) is
about 2.88.
0 Ax0 1 By0 1 C 0
4. a. d 5
"A 2 1 B 2
If you substitute in the coordinates (0, 0), the
0 A(0) 1 B(0) 1 C 0
formula changes to d 5
,
"A 2 1 B 2
0C0
which reduces to d 5
.
"A 2 1 B 2
b. 3x 2 4y 2 12 5 0 and 3x 2 4y 1 12 5 0
0C0
0 212 0
d(L1 ) 5
5
"A 2 1 B 2
"32 1 (24)2
12
5
5
0C0
0 12 0
d(L2 ) 5
5
2
2
2
"A 1 B
"3 1 (24)2
12
5
5
The distance between these parallel lines is
12
12
24
5 1 5 5 5 , because one of the lines is below
the origin and the other is above the origin.
c. 3x 2 4y 2 12 5 0 and 3x 2 4y 1 12 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
Chapter 9: Relationships Between Points, Lines, and Planes
3(0) 2 4y 2 12 5 0 or y 5 23 which corresponds
to the point (0, 3).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(0) 2 4(23) 1 12 0
d5
"32 1 (24)2
24
5
5
Both the answers to 4.b. and 4.c. are the same.
>
5. a. r 5 (22, 1) 1 s(3, 4), sPR
>
r 5 (1, 0) 1 t(3, 4), tPR
First find a random point on one of the lines. We
will use (22, 1) from the first equation. We start by
writing the second equation in parametric form.
Doing so gives x 5 1 1 3t, y 5 4t. We construct a
vector from P(22, 1) to a general point on the line.
>
a 5 322 2 (1 1 3t), 1 2 (4t)4
5 (23 2 3t, 1 2 4t).
(3, 4) ? (23 2 3t, 1 2 4t) 5 0
(29 2 9t) 1 (4 2 16t) 5 0
1
t52
5
This means that the minimal distance between
P(22, 1) and line occurs when t 5 2 15. This point
corresponds to A 25, 2 45 B. The distance between this
point and (22, 1) is 3
x
x21
y
y11
and 5
5
4
23
4
23
First change one equation into a Cartesian equation,
which leads to 3x 1 4y 2 3 5 0 and take a point
from the other equation such as (4, 24).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 3(4) 1 4(24) 2 3 0
d5
"32 1 42
7
5 or 1.4
5
b.
c. 2x 2 3y 1 1 5 0 and 2x 2 3y 2 3 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
2(0) 2 3y 2 3 5 0 or y 5 21 which corresponds
to the point (0, 21).
Calculus and Vectors Solutions Manual
d5
d5
5
0 Ax0 1 By0 1 C 0
"A 2 1 B 2
0 2(0) 2 3(21) 1 1 0
4
"22 1 (23)2
or 1.11
"13
d. 5x 1 12y 5 120 and 5x 1 12y 1 120 5 0
In order to find the distance between these two
parallel lines, you must first find a point on one of
the lines. It is easiest to find a point where the line
crosses the x or y-axis.
5(0) 1 12y 5 120 or y 5 10 which corresponds to
the point (0, 10).
0 Ax0 1 By0 1 C 0
d5
"A 2 1 B 2
0 5(0) 1 12(10) 1 120 0
d5
"52 1 122
240
or 18.46
5
13
>
6. a. P(1, 2, 21) r 5 (1, 0, 0) 1 s(2, 21, 2), sPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 1 1 2s,
y 5 2s, and z 5 2s. We construct a vector from
P(1, 2, 21) to a general point on the line.
>
a 5 31 2 (1 1 2s), 2 2 (2s), 21 2 (2s)4
5 (22s, 2 1 s, 21 2 2s).
(2, 21, 2) ? (22s, 2 1 s, 21 2 2s) 5 0
(24s) 1 (22 2 s) 1 (22 2 4s) 5 0
4
s52
9
This means that the minimal distance between
P(1, 2, 21) and the line occurs when s 5 2 49.
This point corresponds to A 19, 49, 2 89 B. The distance
between this point and P(1, 2, 21) is 1.80.
>
b. P(0, 21, 0) r 5 (2, 1, 0) 1 t(24, 5, 20), tPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 2 2 4t,
y 5 1 1 5t, and z 5 20t. We construct a vector
from P(0, 21, 0) to a general point on the line.
>
a 5 30 2 (2 2 4t), 21 2 (1 1 5t), 0 2 (20t)4
5 (22 1 4t, 22 2 5t, 20t).
(24, 5, 20) ? (22 1 4t, 22 2 5t, 220t) 5 0
(8 2 16t) 1 (210 2 25t) 1 (2400t) 5 0
2
t52
441
This means that the minimal distance between
2
P(0, 21, 0) and the line occurs when t 5 2 441
.
9-21
431
40
This point corresponds to A 890
441 , 441 , 2 441 B. The
distance between this point and P(0, 21, 0) is 2.83.
>
c. P(2, 3, 1) r 5 p(12, 23, 4), pPR
We start by writing the given equation of the line in
parametric form. Doing so gives x 5 12p, y 5 23p,
and z 5 4p. We construct a vector from P(2, 3, 1)
to a general point on the line.
>
a 5 32 2 (12p), 3 2 (23p), 1 2 (4p)4
5 (2 2 12p, 3 1 3p, 1 2 4p).
(12, 23, 4) ? (2 2 12p, 3 1 3p, 1 2 4p) 5 0
(24 2 144p) 1 (29 2 9p) 1 (4 2 16p) 5 0
19
p5
169
This means that the minimal distance between
19
P(2, 3, 1) and the line occurs when p 5 169
. This
57 76
point corresponds to A 228
169 , 2 169 , 169 B. The distance
between this point and P(2, 3, 1) is 3.44.
>
7. a. r 5 (1, 1, 0) 1 s(2, 1, 2), sPR
>
r 5 (21, 1, 2) 1 t(2, 1, 2), tPR
First find a random point on one of the lines. We
will use P(21, 1, 2) from the second equation. We
then write the first equation in parametric form.
Doing so gives x 5 1 1 2s, y 5 1 1 s, and
z 5 0 1 2s. We construct a vector from P(21, 1, 2)
to a general point on the line.
>
a 5 321 2 (1 1 2s), 1 2 (1 1 s), 2 2 2s4
5 (22 2 2s, 2 2 2s).
(2, 1, 2) ? (22 2 2s, 2s, 2 2 2s) 5 0
(24 2 4s) 1 (2s) 1 (4 2 4s) 5 0
s50
This means that the minimal distance between
P(21, 1, 2) and line occurs when s 5 0. This point
corresponds to (1, 1, 0). The distance between this
point and (21, 1, 2) is 2.83
>
b. r 5 (3, 1, 22) 1 m(1, 1, 3), mPR
>
r 5 (1, 0, 1) 1 n(1, 1, 3), nPR
First find a random point on one of the lines.
We will use P(1, 0, 1) from the second equation.
We then write the first equation in parametric form.
Doing so gives x 5 3 1 m, y 5 1 1 m, and
z 5 22 1 3m. We construct a vector from P(1, 0, 1)
to a general point on the line.
>
a 5 31 2 (3 1 m), 0 2 (1 1 m), 1 2 (22 1 3m)4
5 (22 2 3m, 21 2 m, 3 2 3m).
(1, 1, 3) ? (22 2 3m, 21 2 m, 3 2 3m) 5 0
(22 2 3m) 1 (21 2 m) 1 (9 2 9m) 5 0
6
m5
13
9-22
This means that the minimal distance between
P(1, 0, 1) and line occurs when m 5 136 . This point
19
6
corresponds to A 45
13 , 13 , 2 13 B. The distance between
this point and (1, 0, 1) is 3.28
>
8. a. r 5 (1, 21, 2) 1 s(1, 3, 21), sPR
First we write the equation in parametric form.
Doing so gives x 5 1 1 s, y 5 21 1 3s, and
z 5 2 2 s. We construct a vector from P(2, 1, 3) to
a general point on the line.
>
a 5 32 2 (1 1 s), 1 2 (21 1 3s), 3 2 (2 2 s)4
5 (1 2 s, 2 2 3s, 1 1 s).
(1, 3, 21) ? (1 2 s, 2 2 3s, 1 1 s) 5 0
(1 2 s) 1 (6 2 9s) 1 (1 1 s) 5 0
6
s5
11
This means that the minimal distance between
P(2, 1, 3) and line occurs when s 5 116 . This point
7 16
corresponds to A 17
11 , 11 , 11 B.
7 16
b. The distance between A 17
11 , 11 , 11 B and (2, 1, 3)
is 1.65.
9. First, find the line L of intersection between the
planes
1 x 2 y 1 2z 5 2
2
x 1 y 2 z 5 22
Subtract the first equation from the second to
eliminate x and get the equivalent system
1 x 2 y 1 2z 5 2
2
2y 2 3z 5 24
Let z 5 t. Then the second equation gives
2y 5 3t 2 4
3
y5 t22
2
So substituting these into the first equation gives
x 5 y 2 2z 1 2
3
5 a t 2 2b 2 2t 1 2
2
1
52 t
2
So the equation of the line of intersection for these
two planes in parametric form is
1
3
x 5 2 t, y 5 t 2 2, z 5 t, tPR.
2
2
The direction vector for this line is A2 12, 32, 1B,
which is parallel to (21, 3, 2). So, to make things
easier, the parametric form of this line of
intersection could also be expressed as
x 5 2t, y 5 3t 2 2, z 5 2t, tPR
In vector form, this is the same as
Chapter 9: Relationships Between Points, Lines, and Planes
>
r 5 (0, 22, 0) 1 t(21, 3, 2), tPR.
Since Q(0, 22, 0) is on this line,
>
QP 5 (21, 2, 21) 2 (0, 22, 0)
5 (21, 4, 21)
So the distance from P(21, 2, 21) to the line of
intersection is
0 (21, 3, 2) 3 (21, 4, 21) 0
d5
0 (21, 3, 2) 0
0 (211, 23, 21) 0
5
0 (21, 3, 2) 0
131
Å 14
8 3.06
To find the point on the line that gives this minimal
distance, let (x, y, z) be a point on the line. Then,
using the parametric equations,
(x, y, z) 5 (2t, 3t 2 2, 2t)
So the distance from P to this point is
"(x 1 1)2 1 (y 2 2)2 1 (z 1 1)2
5 "(1 2 t)2 1 (3t 2 4)2 1 (2t 1 1)2
5
5 "14t 2 2 22t 1 18
To get the minimal distance, set this quantity
equal to #131
14 .
"14t 2 2 22t 1 18 5
131
Å 14
131
14t 2 2 22t 1 18 5
14
196t 2 2 308t 1 252 5 131
196t 2 2 308t 1 121 5 0
308 6 "0
t5
392
11
5
14
So the point on the line at minimal distance from P is
(x, y, z) 5 (2t, 3t, 22, 2t)
11
11
11
5 a2 , 3a b 2 2, 2a bb
14
14
14
11 5 22
5 a2 , , b
14 14 14
10. A point on the line
>
r 5 (0, 0, 1) 1 s(4, 2, 1), sPR.
has parametric equations
x 5 4s, y 5 2s, z 5 1 1 s, sPR.
Let this point be called
Q(4s, 2s, 1 1 s). Then
>
QA 5 (2, 4, 25) 2 (4s, 2s, 1 1 s)
5 (2 2 4s, 4 2 2s, 26 2 s)
Calculus and Vectors Solutions Manual
If Q is at minimal distance from A, then this vector
will be perpendicular to the direction vector for the
line, (4, 2, 1). This means that
0 5 (2 2 4s, 4 2 2s, 26 2 s) ? (4, 2, 1)
5 10 2 21s
10
s5
21
So the point Q on the line at minimal distance from
A is
10
10
10
Q(4s, 2s, 1 1 s) 5 Qa4a b, 2a b, 1 1 b
21
21
21
40 20 31
5 Qa , , b
21 21 21
Also
>
40
20
31
QA 5 a2 2 , 4 2 , 25 2 b
21
21
21
2 64 136
5 a , ,2
b
21 21
21
So the point Ar will satisfy
>
>
QAr 5 2QA
2
64 136
5 a2 , 2 ,
b
21 21 21
5 Ar(a, b, c) 2 Q
40
20
31
5 aa 2 , b 2 , c 2 b
21
21
21
38
44
167
So a 5 21, b 5 2 21, and c 5 21 . That is,
44 167
Ar( 38
21 , 2 21 , 21 ).
11. a. Think of H as being the origin, E as being on
the x-axis, D as being on the z-axis, and G as being
on the y-axis. That is,
H(0, 0, 0)
E(3, 0, 0)
G(0, 2, 0)
D(0, 0, 2)
and so on for the other points as well. Then line
segment HB has direction vector
B(3, 2, 2) 2 H(0, 0, 0) 5 (3, 2, 2).
>
Also, HA 5 (3, 0, 2). So the distance formula says
that the distance between A and line segment HB is
0 (3, 2, 2) 3 (3, 0, 2) 0
d5
0 (3, 2, 2) 0
0 4, 0, 26 0
5
0 (3, 2, 2) 0
52
Å 17
8 1.75
5
9-23
b. Vertices D and G will give the same distance to
HB because they are equidistant to the segment HB.
(This is easy to check with the distance formula
used similarly to part a. The vertices C, E, and F
give different distances than those found in part a.)
c. The height of triangle AHB was found in part a.,
and was # 52
base length of this triangle is the
17 . The
>
magnitude of HB 5 (3, 2, 2), which is # 52. So
the area of this triangle is
1
52
1
a
b ("17) 5 ("52)
2 Å 17
2
8 3.6 units 2
9.6 The Distance from a Point to a
Plane, pp. 549–550
1. a. Yes the calculations are correct, Point A lies in
the plane.
b. The answer 0 means that the point lies in the
plane.
2. Use the distance formula.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. The distance from A (3, 1, 0) to the plane
20x 2 4y 1 5z 1 7 5 0 is
0 20(3) 1 24(1) 1 5(0) 1 7 0
d5
"202 1 (24)2 1 52
53
b. The distance from B(0, 21, 0) to the plane
2x 1 y 1 2z 2 8 5 0 is
0 2(0) 1 1(21) 1 2(0) 2 8 0
d5
"22 1 12 1 22
53
c. The distance from C(5, 1, 4) to the plane
3x 2 4y 2 1 5 0 is
0 3(5) 1 24(1) 1 0(4) 2 1 0
d5
"32 1 (24)2 1 02
52
d. The distance from D(1, 0, 0) to the plane
5x 2 12y 5 0 is
0 5(1) 2 12(0) 1 0(0) 1 0 0
d5
"5 2 1 (212)2 1 02
5
or 0.38
5
13
9-24
e. The distance from E(21, 0, 1) to the plane
18x 2 9y 1 18z 2 11 5 0 is
0 18(21) 2 9(0) 1 18(1) 2 11 0
d5
"182 1 (29)2 1 182
11
or 0.41
5
27
3. a. 3x 1 4y 2 12z 2 26 5 0 and
3x 1 4y 2 12z 1 39 5 0
First find a point in the second plane such as
(23, 0, 0). Then use d 5
0 Ax0 1 By0 1 Cz0 1 D 0
!A2 1 B2 1 C2
to solve.
0 3(213) 1 4(0) 2 12(0) 2 26 0
d5
"32 1 42 1 (212)2
55
b. 3x 1 4y 2 12z 2 26 5 0
1 3x 1 4y 2 12z 1 39 5 0
6x 1 8y 2 24z 1 13 5 0
c. Answers may vary. Any point on the plane
6x 1 8y 2 24z 1 13 5 0 will work, for example
(2 16, 0, 12) .
4. a. The distance from P(1, 1, 23) to the plane
y 1 3 5 0 is
0 0(1) 1 1(1) 1 0(23) 1 3 0
d5
"02 1 (1)2 1 02
54
b. The distance from Q(21, 1, 4) to the plane
x 2 3 5 0 is
0 1(21) 1 0(1) 1 0(4) 2 3 0
d5
"12 1 02 1 02
54
c. The distance from R(1, 0, 1) to the plane
z 1 1 5 0 is
0 0(1) 1 0(0) 1 1(1) 1 1 0
d5
"02 1 02 1 12
52
5. First you have to find an equation of a plane to
the three points. The equation to this plane is
14x 2 28y 1 28z 2 42 5 0. Then use
d5
0 Ax0 1 By0 1 Cz0 1 D 0
d5
0 14(1) 2 28(21) 1 28(1) 2 42 0
5
!A2 1 B2 1 C2
to solve for the distance.
"142 1 (228)2 1 282
2
or 0.67
3
Chapter 9: Relationships Between Points, Lines, and Planes
b. Vertices D and G will give the same distance to
HB because they are equidistant to the segment HB.
(This is easy to check with the distance formula
used similarly to part a. The vertices C, E, and F
give different distances than those found in part a.)
c. The height of triangle AHB was found in part a.,
and was # 52
base length of this triangle is the
17 . The
>
magnitude of HB 5 (3, 2, 2), which is # 52. So
the area of this triangle is
1
52
1
a
b ("17) 5 ("52)
2 Å 17
2
8 3.6 units 2
9.6 The Distance from a Point to a
Plane, pp. 549–550
1. a. Yes the calculations are correct, Point A lies in
the plane.
b. The answer 0 means that the point lies in the
plane.
2. Use the distance formula.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. The distance from A (3, 1, 0) to the plane
20x 2 4y 1 5z 1 7 5 0 is
0 20(3) 1 24(1) 1 5(0) 1 7 0
d5
"202 1 (24)2 1 52
53
b. The distance from B(0, 21, 0) to the plane
2x 1 y 1 2z 2 8 5 0 is
0 2(0) 1 1(21) 1 2(0) 2 8 0
d5
"22 1 12 1 22
53
c. The distance from C(5, 1, 4) to the plane
3x 2 4y 2 1 5 0 is
0 3(5) 1 24(1) 1 0(4) 2 1 0
d5
"32 1 (24)2 1 02
52
d. The distance from D(1, 0, 0) to the plane
5x 2 12y 5 0 is
0 5(1) 2 12(0) 1 0(0) 1 0 0
d5
"5 2 1 (212)2 1 02
5
or 0.38
5
13
9-24
e. The distance from E(21, 0, 1) to the plane
18x 2 9y 1 18z 2 11 5 0 is
0 18(21) 2 9(0) 1 18(1) 2 11 0
d5
"182 1 (29)2 1 182
11
or 0.41
5
27
3. a. 3x 1 4y 2 12z 2 26 5 0 and
3x 1 4y 2 12z 1 39 5 0
First find a point in the second plane such as
(23, 0, 0). Then use d 5
0 Ax0 1 By0 1 Cz0 1 D 0
!A2 1 B2 1 C2
to solve.
0 3(213) 1 4(0) 2 12(0) 2 26 0
d5
"32 1 42 1 (212)2
55
b. 3x 1 4y 2 12z 2 26 5 0
1 3x 1 4y 2 12z 1 39 5 0
6x 1 8y 2 24z 1 13 5 0
c. Answers may vary. Any point on the plane
6x 1 8y 2 24z 1 13 5 0 will work, for example
(2 16, 0, 12) .
4. a. The distance from P(1, 1, 23) to the plane
y 1 3 5 0 is
0 0(1) 1 1(1) 1 0(23) 1 3 0
d5
"02 1 (1)2 1 02
54
b. The distance from Q(21, 1, 4) to the plane
x 2 3 5 0 is
0 1(21) 1 0(1) 1 0(4) 2 3 0
d5
"12 1 02 1 02
54
c. The distance from R(1, 0, 1) to the plane
z 1 1 5 0 is
0 0(1) 1 0(0) 1 1(1) 1 1 0
d5
"02 1 02 1 12
52
5. First you have to find an equation of a plane to
the three points. The equation to this plane is
14x 2 28y 1 28z 2 42 5 0. Then use
d5
0 Ax0 1 By0 1 Cz0 1 D 0
d5
0 14(1) 2 28(21) 1 28(1) 2 42 0
5
!A2 1 B2 1 C2
to solve for the distance.
"142 1 (228)2 1 282
2
or 0.67
3
Chapter 9: Relationships Between Points, Lines, and Planes
6. 3 5
0 A(3) 2 2(23) 1 6(1) 1 0 0
"A 1 (22) 1 6
3" ( A 1 40 ) 5 Z 3A 1 12 Z
" ( A2 1 40 ) 5 Z A 1 4 Z
A2 1 40 5 A2 1 8A 1 16
24 5 8A
35A
A 5 3 is the only solution to this equation.
7. These lines are skew lines, and the plane containing
>
the second line, r 5 (0, 0, 1) 1 t(1, 1, 0), tPR,
that is parallel to the first line will have direction
vectors (1, 1, 0) and (3, 0, 1). So a normal to this
plane is (1, 1, 0) 3 (3, 0, 1) 5 (1, 21, 23).
So the equation of this plane will be of the form
x 2 y 2 3z 1 D 5 0. We want the point (0, 0, 1)
to be on this plane, and substituting this into the
equation above gives D 5 3. So the equation of the
>
plane containing r 5 (0, 0, 1) 1 t(1, 1, 0), tPR
and parallel to the first line is
x 2 y 2 3z 1 3 5 0.
Since (0, 1, 21) is on the first line, the distance
between these skew lines is the same as the distance
between this point and the plane just determined.
By the distance formula, this distance is
0 (0) 2 (1) 2 3(21) 1 3 0
d5
"12 1 (21)2 1 (23)2
5
5
"11
8 1.51.
8. a. –b. We will do both of these parts at once.
The two given lines are
>
r 5 (1, 22, 5) 1 s(0, 1, 21), sPR,
>
r 5 (1, 21, 22) 1 t(1, 0, 21), tPR.
By converting to parametric form, a general point
on the first line is
U(1, s 2 2, 5 2 s),
and on the second line is
V(1 1 t, 21, 22 2 t).
So the
vector
>
UV 5 (t, 1 2 s, s 2 t 2 7).
If the points U and V are those that produce the >
minimal distance between these two lines, then UV
will be perpendicular to both direction vectors,
(0, 1, 21) and (1, 0, 21). In the first case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (0, 1, 21)
5 8 2 2s 1 t
t 5 2s 2 8
In the second case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (1, 0, 21)
2
2
2
2
Calculus and Vectors Solutions Manual
5 2t 2 s 1 7
Substituting t 5 2s 2 8 into this second equation,
we get
2(2s 2 8) 2 s 1 7 5 0
s53
t 5 2s 2 8
t 5 22
Substituting these values for s and t into U and V,
we get
U(1, 1, 2)
V(21, 21, 0)
So U(1, 1, 2) is the point on the first line that
produces the minimal distance to the second line
at point V(21, 21, 0). This minimal distance is
given by
>
0 UV 0 5 0 (22, 22, 22) 0
5 "12
8 3.46
Review Exercise, pp. 552–555
1. 2x 2 y 5 31, x 1 8y 5 234, 3x 1 ky 5 38
(2x 2 y 5 31) 2 2(x 1 8y 5 234)
5 0x 2 17y 5 99
214
99
y52 ,x5
17
17
214
299
3a
b 1 ka
b 5 38
17
17
4
k52
99
1
2.
x 2 y 5 13
2 3x 1 2y 5 26
3
x 1 2y 5 219
(2 3 Equation 1 ) 1 equation 2 5 5x 1 0y 5 20
or x 5 4. Substituting x 5 4 into equation 1 gives
(4) 2 y 5 13 or y 5 29. However, when you
substitute this coordinates into the third equation,
the third equation is not consistent, so there is no
solution to this problem.
3. a. 1
x 2 y 1 2z 5 3
2 2x 2 2y 1 3z 5 1
3
2x 2 2y 1 z 5 11
Equation 2 2 equation 3 5 5z 5 210 or
z 5 22. Substituting z 5 22 into all of the equations
gives
4
x2y2453
5 2x 2 2y 2 6 5 1
6 2x 2 2y 2 2 5 11
There are no x and y variables that satisfy these
equations, so the answer is no solution.
9-25
6. 3 5
0 A(3) 2 2(23) 1 6(1) 1 0 0
"A 1 (22) 1 6
3" ( A 1 40 ) 5 Z 3A 1 12 Z
" ( A2 1 40 ) 5 Z A 1 4 Z
A2 1 40 5 A2 1 8A 1 16
24 5 8A
35A
A 5 3 is the only solution to this equation.
7. These lines are skew lines, and the plane containing
>
the second line, r 5 (0, 0, 1) 1 t(1, 1, 0), tPR,
that is parallel to the first line will have direction
vectors (1, 1, 0) and (3, 0, 1). So a normal to this
plane is (1, 1, 0) 3 (3, 0, 1) 5 (1, 21, 23).
So the equation of this plane will be of the form
x 2 y 2 3z 1 D 5 0. We want the point (0, 0, 1)
to be on this plane, and substituting this into the
equation above gives D 5 3. So the equation of the
>
plane containing r 5 (0, 0, 1) 1 t(1, 1, 0), tPR
and parallel to the first line is
x 2 y 2 3z 1 3 5 0.
Since (0, 1, 21) is on the first line, the distance
between these skew lines is the same as the distance
between this point and the plane just determined.
By the distance formula, this distance is
0 (0) 2 (1) 2 3(21) 1 3 0
d5
"12 1 (21)2 1 (23)2
5
5
"11
8 1.51.
8. a. –b. We will do both of these parts at once.
The two given lines are
>
r 5 (1, 22, 5) 1 s(0, 1, 21), sPR,
>
r 5 (1, 21, 22) 1 t(1, 0, 21), tPR.
By converting to parametric form, a general point
on the first line is
U(1, s 2 2, 5 2 s),
and on the second line is
V(1 1 t, 21, 22 2 t).
So the
vector
>
UV 5 (t, 1 2 s, s 2 t 2 7).
If the points U and V are those that produce the >
minimal distance between these two lines, then UV
will be perpendicular to both direction vectors,
(0, 1, 21) and (1, 0, 21). In the first case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (0, 1, 21)
5 8 2 2s 1 t
t 5 2s 2 8
In the second case, we get
0 5 (t, 1 2 s, s 2 t 2 7) ? (1, 0, 21)
2
2
2
2
Calculus and Vectors Solutions Manual
5 2t 2 s 1 7
Substituting t 5 2s 2 8 into this second equation,
we get
2(2s 2 8) 2 s 1 7 5 0
s53
t 5 2s 2 8
t 5 22
Substituting these values for s and t into U and V,
we get
U(1, 1, 2)
V(21, 21, 0)
So U(1, 1, 2) is the point on the first line that
produces the minimal distance to the second line
at point V(21, 21, 0). This minimal distance is
given by
>
0 UV 0 5 0 (22, 22, 22) 0
5 "12
8 3.46
Review Exercise, pp. 552–555
1. 2x 2 y 5 31, x 1 8y 5 234, 3x 1 ky 5 38
(2x 2 y 5 31) 2 2(x 1 8y 5 234)
5 0x 2 17y 5 99
214
99
y52 ,x5
17
17
214
299
3a
b 1 ka
b 5 38
17
17
4
k52
99
1
2.
x 2 y 5 13
2 3x 1 2y 5 26
3
x 1 2y 5 219
(2 3 Equation 1 ) 1 equation 2 5 5x 1 0y 5 20
or x 5 4. Substituting x 5 4 into equation 1 gives
(4) 2 y 5 13 or y 5 29. However, when you
substitute this coordinates into the third equation,
the third equation is not consistent, so there is no
solution to this problem.
3. a. 1
x 2 y 1 2z 5 3
2 2x 2 2y 1 3z 5 1
3
2x 2 2y 1 z 5 11
Equation 2 2 equation 3 5 5z 5 210 or
z 5 22. Substituting z 5 22 into all of the equations
gives
4
x2y2453
5 2x 2 2y 2 6 5 1
6 2x 2 2y 2 2 5 11
There are no x and y variables that satisfy these
equations, so the answer is no solution.
9-25
x 1 y 1 z 5 300
x 1 y 2 z 5 98
3 x 2 y 1 z 5 100
Equation 2 1 equation 3 5 2x 5 198 x 5 99.
Substituting x 5 99 into all three equations gives:
4
y 1 z 5 201
5
y 2 z 5 21
6 2y 1 z 5 1
Equation 4 1 equation 5 5 2y 5 200 or
y 5 100. You then get z 5 101 after substituting
both x and y into equation 1 .
(99, 100, 101)
Check:
1 99 1 100 1 101 5 300
2 99 1 100 2 101 5 98
3 99 2 100 1 101 5 100
4. a. These four points will lie in the same plane if
and only if the line determined by the first two
points intersects the line determined by the last two
points. The direction vector determined by the first
two is
>
a 5 (7, 25, 1) 2 (1, 2, 6)
5 (6, 27, 25)
So these first two points determine the line with
vector equation
>
r 5 (1, 2, 6) 1 s(6,27,25), sPR.
The direction vector determined by the last two
points is
>
b 5 (23, 5, 6) 2 (1, 1, 4)
5 (24, 4, 2)
So these first two points determine the line with
vector equation
>
r 5 (1, 1, 4) 1 t(24, 4, 2), tPR.
Converting these two lines to parametric form, we
obtain the equations
1
1 1 6s 5 1 2 4t
2
2 2 7s 5 1 1 4t
3
6 2 5s 5 4 1 2t
Adding the first and second equations gives
3 2 s 5 2, so s 5 1. Substituting this into the third
equation, we get
1 5 4 1 2t
23 5 2t
So t 5 2 32. We need to check this s and t for
consistency. Substituting s 5 1 into the vector
equation for the first line gives
>
r 5 (1, 2, 6) 1 (1)(6, 27, 25)
5 (7, 25, 1)
as a point on this line. Substituting t 5 2 32 into the
vector equation for the second line gives
b.
1
2
9-26
3
>
r 5 (1, 1, 4) 1 a2 b (24, 4, 2)
2
5 (1, 1, 4) 1 (6, 26, 23)
5 (7, 25, 1)
as a point on this line. This means the two lines
intersect, and so the four points given lie in the
same plane.
b. Direction vectors for the plane containing the
four points in part a. are (6, 27, 25) and
(24, 4, 2). So a normal to this plane is
(6, 27, 25) 3 (24, 4, 2) 5 (6, 8, 24).
We will use the parallel normal (3, 4, 22). So the
equation of this plane is of the form
3x 1 4y 2 2z 1 D 5 0.
Substitute in the point (1, 2, 6) to find D.
3(1) 1 4(2) 2 2(6) 1 D 5 0
D51
The equation of the plane is
3x 1 4y 2 2z 1 1 5 0.
So, using the distance formula, this plane is distance
Z 3(0) 1 4(0) 2 2(0) 1 1 Z
d5
Z (3, 4, 22) Z
1
5
"29
8 0.19
from the origin.
5. Use the distance formula.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. The distance from A(21, 1, 2) to
3x 2 4y 2 12z 2 8 5 0
0 3(21) 2 4(1) 2 12(2) 2 8 0
d5
"32 1 (24)2 1 (212)2
53
b. The distance from B(3, 1, 22) to
8x 2 8y 1 4z 2 7 5 0
0 8(3) 2 8(1) 1 4(22) 2 7 0
d5
"82 1 (28)2 1 (4)2
1
or 0.08
5
12
>
6. r 5 (3, 1, 1) 1 t(2, 21, 2), tPR
3x 2 4y 2 5z 5 0
Find the parametric equations from the first
equation, then substitute those equations into the
second equation. Solve for t. Substitute that t-value
into the first equation.
>
r 5 (3, 1, 1) 1 t(2, 21, 2), tPR
x 5 3 1 2t
Chapter 9: Relationships Between Points, Lines, and Planes
y512t
z 5 1 1 2t
3(3 1 2t) 2 4(1 2 t) 2 5(1 1 2t) 5 0
t can be any value to satisfy this value, so the two
equations intersect along
>
r 5 (3, 1, 1) 1 t(2, 21, 2), tPR.
7. a. 1
3x 2 4y 1 5z 5 9
2
6x 2 9y 1 10z 5 9
3 9x 2 12y 1 15z 5 9
3 3 (3x 2 4y 1 5z 5 9) 5 9x 2 12y 1 15z 5 27
There is no solution because the first and third
equations are inconsistent.
1 2x 1 3y 1 4z 5 3
b.
2 4x 1 6y 1 8z 5 4
3
5x 1 y 2 z 5 1
2 3 (2x 1 3y 1 4z 5 3) 5 4x 1 6y 1 8z 5 6
There is no solution because the first and second
equations are inconsistent.
1
c.
4x 2 3y 1 2z 5 2
2
8x 2 6y 1 4z 5 4
3 12x 2 9y 1 6z 5 1
3 3 (4x 2 3y 1 2z 5 2) 5 12x 2 9y 1 6z 5 6
There is no solution because the first and third
equations are inconsistent.
8. a. 1 3x 1 4y 1 z 5 4
2 5x 1 2y 1 3z 5 2
3 6x 1 8y 1 2z 5 8
(Equation 1 ) 2 (2 3 equation 2 )
5 27x 2 5z 5 0
Letting z 5 t, then x 5 2 57 t and y 5 1 1 27 t.
5
2
x 5 2 t, y 5 1 1 t, z 5 t, tPR
7
7
1 4x 2 8y 1 12z 5 4
b.
2
2x 1 4y 1 6z 5 4
3
x 2 2y 2 3z 5 4
(Equation 1 ) 1 (4 3 equation 3 )
5 24z 5 212 or z 5 2 12. Letting z 5 2 12 creates:
4x 2 8y 5 10
2x 1 4y 5 7
(Equation 1 ) 1 (2 3 equation 2 ) 5 8x 5 24
or x 5 3. Substituting in x 5 3 and z 5 2 12 gives
y 5 14
1
1
x 5 3, y 5 , z 5 2
4
2
1
c.
x 2 3y 1 3z 5 7
2
2x 2 6y 1 6z 5 14
3 2x 1 3y 2 3z 5 27
4
5
Calculus and Vectors Solutions Manual
Letting z 5 s, then y 5 t gives x 2 3t 1 3s 5 7 or
x 5 23s 1 3t 1 7
x 5 3t 2 3s 1 7, y 5 t, z 5 s, s, tPR
9. a. 1 3x 2 5y 1 2z 5 4
1
6x 1 2y 2 z 5 2
1 6x 2 3y 1 8z 5 6
(Equation 2 ) 2 (2 3 equation 1 ) 5 12y 2 5z
5 26
Setting z 5 t,
1
5
12y 2 5t 5 26 or y 5 2 1 t
2
12
Substituting these two values into the first equation
gives x 5 12 1 361 t
1
1
21
5
x 5 1 t, y 5
1 t, z 5 t, tPR
2
36
2
12
1 2x 2 5y 1 3z 5 1
b.
2 4x 1 2y 1 5z 5 5
3 2x 1 7y 1 2z 5 4
(Equation 2 ) 2 (2 3 equation 1 )
5 12y 2 z 5 3
Setting z 5 t,
1
1
12y 2 t 5 3 or y 5 1 t
4
12
Substituting these two values into the first equation
gives x 5 98 2 31
24 t
1
9
31
1
x 5 2 t, y 5 1 t, z 5 t, tPR
8
24
4
12
10. a. 2x 1 y 1 z 5 6
x 2 y 2 z 5 29
3x 1 y 5 2
The first equation 1 the second equation gives
3x 5 23 or x 5 21. Substituting x 5 21 into the
third equation, 3(21) 1 y 5 2 or y 5 5.
Substituting these two values into the first equation,
2(21) 1 5 1 z 5 6 or z 5 3
These three planes meet at the point (21, 5, 3).
1 2x 2 y 1 2z 5 2
b.
2
3x 1 y 2 z 5 1
3 x 2 3y 1 5z 5 4
Equation 1 1 equation 2 5 5x 1 z 5 3
Equation 3 2 (3 3 equation 1 ) 5 25x 2 z
5 22.
These two equations are inconsistent, so the planes
do not intersect at any point. Geometrically the
planes form a triangular prism.
1
c.
2x 1 y 2 z 5 0
2 x 2 2y 1 3z 5 0
3 9x 1 2y 2 z 5 0
9-27
2 3 equation 1 1 equation 2 5 5x 1 z 5 0, so
z 5 25x.
Equation 3 2 equation 1 5 7x 1 y 5 0, so
y 5 27x.
Let x 5 t. The intersection of the planes is a line
through the origin with equation x 5 t, y 5 27t,
z 5 25t, tPR.
>
11. r 5 (2, 21, 22) 1 s(1, 1, 22), sPR
By substituting in different s-values, you can find
when the plane intersects the xz-plane when y 5 0
and the xy-plane when z 5 0.
The plane intersects the xz-plane at (3, 0, 24) and
the xy-plane at (1, 22, 0). Then find the distance
between these two points using the distance
formula. The distance between these two points
is 4.90.
12. a. x 2 2y 1 z 1 4 5 0
>
r 5 (3, 1, 25) 1 s(2, 1, 0), sPR
> >
m ? n 5 (2, 1, 0) ? (1, 22, 1) 5 0 Since the line’s
direction vector is perpendicular to the normal of
the plane and the point (3, 1, 25) lies on both the
line and the plane, the line is in the plane.
>
b. r 5 (7, 5, 21) 1 t(4, 3, 2), tPR
>
r 5 (3, 1, 25) 1 s(2, 1, 0), sPR
Solve for the parametric equations of both equations
and then set them equal to each other.
L1: x 5 7 1 4t, y 5 5 1 3t, z 5 21 1 2t
L2: x 5 3 1 2s, y 5 1 1 s, z 5 25
z 5 25 5 21 1 2t, t 5 22
t 5 22, x 5 21, y 5 21, z 5 25
t 5 22 corresponds to the point (21, 21, 25)
c.
x 2 2y 1 z 1 4 5 0
21 2 2(21) 1 (25) 1 4 5 0
The point (21, 21, 25) is on the plane since it
satisfies the equation of the plane.
>
d. r 5 (7, 5, 21) 1 t(4, 3, 2), tPR
(A, B, C) ? (4, 3, 2) 5 0
A 5 7, B 5 22, C 5 211
7x 2 2y 2 11z 1 D 5 0
D 5 250
7x 2 2y 2 11z 2 50 5 0
>
13. a. r 5 (3, 0, 21) 1 t(1, 1, 2), tPR
A(22, 1, 1)
x 5 3 1 t, y 5 t, z 5 21 1 2t
0 5 3 1 t 2 x, 0 5 t 2 y, 0 5 21 1 2t 2 z
"(3 1 t 2 x)2 1 (t 2 y)2 1 (21 1 2t 2 z)2
"(3 1 t 1 2)2 1 (t 2 1)2 1 (21 1 2t 2 1)2
"6t 2 1 30
t 5 0 gives the lowest distance of 5.48
9-28
b. t 5 0 corresponds to the point (3, 0, 21)
>
14. a. r 5 (1, 21, 1) 1 t(3, 2, 1), tPR
>
r 5 (22, 23, 0) 1 s(1, 2, 3), sPR
Set the equations parametric equations equal to
each other, and determine either the s or t-value.
Find the point that corresponds to this value.
L1: x 5 1 1 3t, y 5 21 1 2t, z 5 1 1 t
L2: x 5 22 1 s, y 5 23 1 2s, z 5 3s
x 5 1 1 3t 5 22 1 s
y 5 21 1 2t 5 23 1 2s
z 5 1 1 t 5 3s
s 5 0, t 5 21
s 5 0 corresponds to the point (22, 23, 0).
>
b. r 5 (1, 21, 1) 1 t(3, 2, 1), tPR
>
r 5 (22, 23, 0) 1 s(1, 2, 3), sPR
P(22, 23, 0)
S
S
n1 3 n2 5 (3, 2, 1) 3 (1, 2, 3)
5 (4, 28, 4) 5 (1, 22, 1)
>
r 5 (22, 23, 0) 1 t(1, 22, 1), tPR
15. a. Since the plane we want contains L, we can
use the direction vector for L, (1, 2, 21), as one
of the plane’s direction vectors. Since the plane
contains the point (1, 2, 23) (which is on L) and the
point K(3, 22, 4), it will contain the direction vector
(3, 22, 4) 2 (1, 2, 23) 5 (2, 24, 7)
To find a normal vector for the plane we want, take
the cross product of these two direction vectors.
(2, 24, 7) 3 (1, 2, 21) 5 (210, 9, 8)
So the plane we seek will be of the form
210x 1 9y 1 8z 1 D 5 0.
To determine the value of D, substitute in the point
(1, 2, 23) that is to be on this plane.
210(1) 1 9(2) 1 8(23) 1 D 5 0
D 5 16
The equation of the plane we seek is
210x 1 9y 1 8z 1 16 5 0.
b. Using the distance formula, the distance from
S(1, 1, 21) to the plane 210x 1 9y 1 8z 1
16 5 0 is
Z210(1) 1 9(1) 1 8(21) 1 16Z
d5
Z (210, 9, 8)Z
7
5
"245
8 0.45
16. a. 1
x1y2z51
2 2x 2 5y 1 z 5 21
3 7x 2 7y 2 z 5 k
Equation 1 1 equation 2 5 equation 4
5 3x 2 4y 5 0
Chapter 9: Relationships Between Points, Lines, and Planes
Equation 2 1 equation 3 5 equation 5
5 9x 2 12y 5 21 1 k
For the solution to this system to be a line,
equation 4 and equation 5 must be the proportional.
k 5 1 makes these two line proportional and the
solution to this system a line.
b. In part a., we found that k 5 1 by arriving at the
equivalent system
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
3
9x 2 12y 5 0
As the first and third equations are proportional,
this is really the same system as
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
Letting x 5 t in the first equation, we see that
y 5 34 t. Substituting these values for x and y into the
second equation, we find that
3
z 5 5a tb 2 2t 2 1
4
7
5 t 2 1.
4
So the direction vector for the line that solves this
system is (1, 34, 74), which is parallel to (4, 3, 7).
So equivalent parametric equations of this line are
x 5 4t
y 5 3t
z 5 21 1 7t, tPR.
So one possible vector equation of this line is
>
r 5 (0, 0, 21) 1 t(4, 3, 7), tPR.
b. In part a., we found that k 5 1 by arriving at the
equivalent system
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
3
9x 2 12y 5 0
As the first and third equations are proportional,
this is really the same system as
1
3x 2 4y 5 0
2 2x 2 5y 1 z 5 21
Letting x 5 t in the first equation, we see that
y 5 34t. Substituting these values for x and y into the
second equation, we find that
3
z 5 5a tb 2 2t 2 1
4
7
5 t 2 1.
4
So the direction vector for the line that solves this
system is Q 1, 34, 74 R , which is parallel to (4, 3, 7).
So equivalent parametric equations of this line are
Calculus and Vectors Solutions Manual
x 5 4t
y 5 3t
z 5 21 1 7t, tPR.
So one possible vector equation of this line is
>
r 5 (0, 0, 21) 1 t(4, 3, 7), tPR.
17. a. 1
x 1 2y 1 z 5 1
2
2x 2 3y 2 z 5 6
3 3x 1 5y 1 4z 5 5
4
4x 1 y 1 z 5 8
Equation 1 1 equation 2 5 equation 5
5 3x 2 y 5 7
(4 3 equation 2 ) 1 equation 3 5 equation 6
5 11x 2 7y 5 29
(7 3 equation 5 ) 1 equation 6
5 equation 7 5 210x 5 220y or x 5 2
Substituting into equation 5 : 6 2 y 5 7y 5 21.
Substituting into equation 1 : 2 1 22 1 z 5 1
or z 5 1.
(2, 21, 1)
1
b.
x 2 2y 1 z 5 1
2
2x 2 5y 1 z 5 21
3
3x 2 7y 1 2z 5 0
4 6x 2 14y 1 4z 5 0
Equation 2 2 (2 3 equation 1 )
5 equation 5 5 2y 2 z 5 23,
Setting z 5 t,
2y 2 t 5 23 or y 5 3 2 t
Substituting y 5 3 2 t and z 5 t into equation 1 :
x 2 2(3 2 t) 1 t 5 1 or x 5 7 2 3t
x 5 7 2 3t, y 5 3 2 t, z 5 t, tPR
9a
3c
18. 1
2 8b 1
54
b
b
3a
4c
1 4b 1
53
b
b
3a
4c
3
1 4b 2
53
b
b
a
c
x 5 , y 5 b, z 5
b
b
1
9x 2 8y 1 3z 5 4
2 23x 1 4y 1 4z 5 3
3
3x 1 4y 2 4z 5 3
3 1 2 5 8y 5 6
3
y5
4
1 9x 1 3z 5 10
2 23x 1 4z 5 0
3 3x 2 4z 5 0
1 1 3 2 5 15z 5 10
2
2
9-29
8
2
z5 ,x5
3
9
3
y 5 5 b,
4
2
8
a
a
x5 5 5 ,a5
9
b
3
3
4
1
2
c
c
z5 5 5 ,c5
3
b
3
2
4
2 3 1
a , , b
3 4 2
19. First put the equation into parametric form.
Then substitute the x, y, and z-values into
x 1 2y 2 z 1 10 5 0 to determine t. Then
substitute t back into the parametric equations
to determine the coordinates.
x11
y22
z21
5
5
5t
24
3
22
x 5 24t 2 1, y 5 3t 1 2, z 5 22t 1 1
x 1 2y 2 3z 1 10 5 0
(24t 2 1) 1 2(3t 1 2) 2 3(22t 1 1) 1 10 5 0
5
t52
4
5
5
x 5 24a2 b 2 1, y 5 3a2 b 1 2,
4
4
5
z 5 22a2 b 1 1
4
7 7
a4, 2 , b
4 2
20. Let Ar(a, b, c) denote the image point under this
reflection. We want to find a, b, and c. The equation
of the plane is x 2 y 1 z 2 1 5 0, so letting y 5 s
and z 5 t, we get x 5 1 2 t 1 s, s, tPR. These are
the parametric equations of this plane, so a general
point on this plane has coordinates P(1 2 t 1 s, s, t).
>
So PA 5 (1, 0, 4) 2 (1 2 t 1 s, s, t)
5 (t 2 s, 2s, 4 2 t)
The normal vector to this plane is (1, 21, 1), and in
>
order for PA to be perpendicular to the plane, it >
must be parallel to this normal. This means that PA
and (1, 21, 1) will have a cross product equal to
the zero vector.
(t 2 s, 2s, 4 2 t) 3 (1, 21, 1)
5 (4 2 s 2 t, 4 1 s 2 2t, 2s 2 t)
5 (0, 0, 0)
9-30
So we get the system of equations
1
42s2t50
2 4 1 s 2 2t 5 0
3
2s 2 t 5 0
Adding the first two equations gives
8 2 3t 5 0
8
t5
3
Substituting this value for t into the third equation
gives
0 5 2s 2 t
8
5 2s 2
3
4
s5
3
Substituting these values for s and t into the equation
>
for PA, we get
>
PA 5 (t 2 s, 2s, 4 2 t)
8
4 4
8
5 a 2 ,2 ,42 b
3
3 3
3
4 4 4
5 a ,2 , b
3 3 3
This is the vector that is normal to the plane, with
its head at point A(1, 0, 4) and tail at the point in
the plane
8
4 4 8
P(1 2 t 1 s, s, t) 5 Pa1 2 1 , , b
3
3 3 3
1 4 8
5 a2 , , b
3 3 3
So the
vector
>
>
PAr 5 2PA
4 4 4
5 a2 , , 2 b
3 3 3
1 4 8
5 (a, b, c) 2 a2 , , b
3 3 3
1
4
8
5 aa 1 , b 2 , c 2 b
3
3
3
This means that a 5 2 53, b 5 2 83, and c 5 2 43,
That is, the reflected point is Ar( 2 53, 83, 43) .
21. a. The first plane has normal (3, 1, 7) and the
second has normal (4, 212, 4). Their line of
intersection will be perpendicular to both of these
normals. So we can take as direction vector the
cross product of these two normals.
(3, 1, 7) 3 (4, 212, 4) 5 (88, 16, 240)
5 8(11, 2, 25)
Chapter 9: Relationships Between Points, Lines, and Planes
So let’s use (11, 2, 25) as the direction vector for
this line of intersection. To find a point on both of
these planes, solve for z in the second plane, and
substitute this into the equation for the first plane.
4x 2 12y 1 4z 2 24 5 0
4z 5 24 2 4x 1 12y
z 5 6 2 x 1 3y
0 5 3x 1 y 1 7z 1 3
5 3x 1 y 1 7(6 2 x 1 3y)
13
5 24x 1 22y 1 45
If y 5 0 in this last equation, then x 5 454 and
z 5 6 2 x 1 3y
45
562
1 3(0)
4
21
52
4
The point ( 454, 0, 214) , lies on both planes. So the
vector equation of the line of intersection for the
first two planes is
21
45
>
r 5 a , 0, 2 b 1 t(11, 2, 25), tPR.
4
4
The corresponding parametric form is
45
x5
1 11t
4
y 5 2t
21
z 5 2 2 5t, tPR.
4
We will use a similar procedure for the other two
lines of intersection. For the third plane, the normal
vector is (1, 2, 3). So a direction vector for the line
of intersection between the first and third planes is
(3, 1, 7) 3 (1, 2, 3) 5 (211, 22, 5)
5 2 (11, 2, 25)
We may use (11, 2, 25) as the direction vector for
this line of intersection. We find a point on both of
these planes in the same way as before.
x 1 2y 1 3z 2 4 5 0
x 5 4 2 2y 2 3z
0 5 3x 1 y 1 7z 1 3
5 3(4 2 2y 2 3z) 1 y 1 7z 1 3
5 26y 2 2z 1 15
Taking y 5 0 in this last equation, we get z 5 152
and
x 5 4 2 2y 2 3z
15
5 4 2 2(0) 2 3a b
2
37
52
2
Calculus and Vectors Solutions Manual
A point on both the first and third planes is
(2 372, 0, 152 ). So the vector equation for this line of
intersection is
15
37
>
r 5 a2 , 0, b 1 t(11, 2, 25), tPR,
2
2
and the corresponding parametric equations are
37
x 5 2 1 11t
2
y 5 2t
15
z5
2 5t, tPR.
2
Finally, we consider the line of intersection between
the second and third planes. In this case, a direction
vector is
(4, 212, 4) 3 (1, 2, 3) 5 (244, 28, 20)
5 24(11, 2, 25)
We may use (11, 2, 25) as the direction vector for
this line of intersection. We find a point on both of
these planes in the same way as before.
x 1 2y 1 3z 2 4 5 0
x 5 4 2 2y 2 3z
0 5 4x 2 12y 1 4z 2 24
5 4(4 2 2y 2 3z) 2 12y 1 4z 2 24
5 220y 2 8z 2 8
Taking y 5 0 in this last equation, we get z 5 21
and
x 5 4 2 2y 2 3z
5 4 2 2(0) 2 3(21)
57
A point on both the second and third planes is
(7, 0, 21). So the vector equation for this line of
intersection is
>
r 5 (7, 0, 21) 1 t(11, 2, 25), tPR,
and the corresponding parametric equations are
x 5 7 1 11t
y 5 2t
z 5 21 2 5t, PR.
b. All three lines of intersection found in part a.
have direction vector (11, 2, 25), and so they are
all parallel. Since no pair of normal vectors for
these three planes is parallel, no pair of these planes
is coincident.
2
5
3
22. 1 2 1 2 1 2 5 40
a
b
c
3
6
1
2
2 2 2 2 5 23
a2
b
c
5
4
9
3
2 2 1 2 5 67
a2
b
c
9-31
11 213
1 2 5 31
a2
b
21 229
3 14 2 5 5 5
1 2 5 55
a2
b
46
21 4 2 11 5 5 2 5 46, b 5 11, b 5 21
b
1
21
229
1
5 55, a 5 , a 5 2
2 1
a
1
2
2
1
2
5
3
1
1 1 2 5 40, c 5 , c 5 2
0.25
1
c
3
3
1
1
1
1
a 5 , a 5 2 , b 5 1, b 5 21, c 5 , c 5 2
2
2
3
3
2
2
Because each equation has each of a , b , and c2, the
possible solutions are all combinations of the
positive and negative values for a, b, and c: ( 12, 1, 13) ,
1
13
2
5
4
5
( 12, 1, 2 13) , ( 12, 21, 13) , ( 12, 21, 2 13) , ( 2 12, 1, 13) ,
( 2 12, 1, 2 13) , ( 2 12, 21, 13) , and ( 12, 21, 2 13) .
23. The general form of such a parabola is
y 5 ax 2 1 bx 1 c. We need to determine a, b, and c.
Since (21, 2), (1, 21), and (2, 1) all lie on the
parabola, we get the system of equations
1
a2b1c52
2
a 1 b 1 c 5 21
3 4a 1 2b 1 c 5 1
Adding the first and second equations gives
1
a1c5
2
Subtracting the first from the second equation gives
2b 5 23
3
b52
2
Using the fact that a 1 c 5 12 and b 5 2 32 in the
third equation gives
1 5 4a 1 2b 1 c
5 3a 1 2b 1 (a 1 c)
3
1
5 3a 1 2a2 b 1
2
2
5
5 3a 2
2
7
5 3a
2
7
a5
6
So using once more that a 1 c 5 12, we substitute
this value in for a and get
1
5a1c
2
7
1c
6
2
c52
3
So the equation of the parabola we seek is
7
3
2
y 5 x2 2 x 2 .
6
2
3
24. The equation of the plane is
4x 2 5y 1 z 2 9 5 0, which has normal (4, 25, 1).
Converting this plane to parametric form gives
x5s
y5t
z 5 9 2 4s 1 5t, s, tPR.
So for any point Y(s, t, 9 2 4s 1 5t) on this plane,
we >can form the vector
XY 5 (s, t, 9 2 4s 1 5t) 2 (3, 2, 25)
5 (s 2 3, t 2 2, 14 2 4s 1 5t)
This vector is perpendicular to the plane when it
is parallel to the normal vector (4, 25, 1). Two
vectors are parallel precisely when their cross
product is the zero vector.
(s 2 3, t 2 2, 14 2 4s 1 5t) 3 (4, 25, 1)
5 (68 1 26t 2 20s, 59 1 20t 2 17s, 23 2 4t 2 5s)
5 (0, 0, 0)
So we get the system of equations
1 68 1 26t 2 20s 5 0
2 59 1 20t 2 17s 5 0
3
23 2 4t 2 5s 5 0
Subtracting four times the third equation from the
first equation gives
42t 2 24 5 0
4
t5
7
Substituting this value for t into the second equation
gives
0 5 59 1 20t 2 17s
4
5 59 1 20a b 2 17s
7
493
17s 5
7
29
s5
7
Substituting these values for s and t into the equation
for Y gives
29
29 4
Y(s, t, 9 2 4s 1 5t) 5 Ya , , 9 2 4a b
7 7
7
4
29 4 33
1 5a bb 5 a , , 2 b
7
7 7
7
5
So the point M we wanted is M( 297, 47, 2 337 ).
9-32
Chapter 9: Relationships Between Points, Lines, and Planes
11x 2 2 14x 1 9
A
Bx 1 C
5
1 2
2
(3x 2 1)(x 1 1)
3x 2 1
x 11
11x 2 2 14x 1 9
(3x 2 1)(x 2 1 1)
A(x 2 1 1) 1 (Bx 1 C) 3x 2 1
5
(3x 2 1)(x 2 1 1)
11x 2 2 14x 1 9 5 (A 1 3B)x 2 1 (3C 2 B)x
1 (A 2 C)
A 2 C 5 9, 3C 2 B 5 214, A 1 3B 5 11
B 5 3C 1 14, A 5 C 1 9
A 1 3(3C 1 14) 5 11, A 1 9C 5 231
(C 1 9) 1 9C 5 231
10C 5 240, C 5 24
B 5 3(24) 1 14 5 2, A 5 (24) 1 9 5 5
A 5 5, B 5 2, C 5 24
26.> a. The vector
EF 5 (21, 24, 26) 2 (4, 0, 3)
5 (25, 24, 23)
This is a direction vector for the line containing the
segment EF. The point E(21, 24, 26) is on this
line, so the vector equation of this line is
>
r 5 (21, 24, 26) 1 t(25, 24, 23), tPR.
b. Based on the equation of the line found in part a.,
a general point on this line is of the form
J(21 2 5t, 24 2 4t, 26 2 3t), tPR.
For> this general point, the vector
JD 5 (3, 0, 7) 2 (21 2 5t, 24 24t, 26 2 3t)
5 (4 1 5t, 4 1 4t, 13 1 3t)
This vector will be perpendicular to the direction
vector for the line found in part a. at the point J we
seek. This means that
0 5 (4 1 5t, 4 1 4t, 13 1 3t) ? (25, 24, 23)
5 25(4 1 5t) 2 4(4 1 4t) 2 3(13 1 3t)
5 275 2 50t
3
t52
2
Substituting this value of t into the equation for the
general point on the line in part a.,
J(21 2 5t, 24 2 4t, 26 2 3t)
3
3
3
5 Ja21 2 5a2 b, 24 2 4a2 b, 26 2 3a2 bb
2
2
2
13
3
5 a , 2, 2 b
2
2
These are the coordinates for the point J we wanted.
c. Using the coordinates for J found in part b.,
>
13
3
JD 5 (3, 0, 7) 2 a , 2, 2 b
2
2
7
17
5 a2 , 22, b
2
2
25.
Calculus and Vectors Solutions Manual
This vector forms the height of ^ DEF, and the
length of this vector is
>
7
17
@ JD @ 5 ` a2 , 22, b `
2
2
2
7
17 2
5 a2 b 1 (22)2 1 a b
Å
2
2
177
Å 72
8 9.41
The length of the base of ^ DEF is
>
@ EF @ 5 0 (25, 24, 23) 0
5
5 "(25)2 1 (24)2 1 (23)2
5 "50
8 7.07
So the area of ^ DEF equals
177
5
1
( !50) a
b 5 !177
2
Å 2
2
8 33.26 units2
27. 3x 2 2z 1 1 5 0
4x 1 3y 1 7 5 0
(5, 25, 5)
>
>
n1 3 n2 5 (3, 0, 22) 3 (4, 3, 0) 5 (6, 28, 9)
6x 2 8y 1 9z 1 D 5 0
D 5 2115
6x 2 8y 1 9z 2 115 5 0
Chapter 9 Test, p. 556
>
1. a. r 1 5 (4, 2, 6) 1 s(1, 3, 11), sPR,
>
r 2 5 (5, 21, 4) 1 t(2, 0, 9), tPR
L1: x 5 4 1 s, y 5 2 1 3s, z 5 6 1 11s
L2: x 5 5 1 2t, y 5 21, z 5 4 1 9t
y 5 21 5 2 1 3s
s 5 21
L1: x 5 4 1 (21), y 5 2 1 3(21),
z 5 6 1 11(21)
x 5 3, y 5 21, z 5 25
(3, 21, 25)
b.
x2y1z1150
3 2 (21) 1 (25) 1 1 5 0
311251150
050
2. Use the distance equation.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. A(3, 2, 3)
8x 2 8y 1 4z 2 7 5 0
9-33
11x 2 2 14x 1 9
A
Bx 1 C
5
1 2
2
(3x 2 1)(x 1 1)
3x 2 1
x 11
11x 2 2 14x 1 9
(3x 2 1)(x 2 1 1)
A(x 2 1 1) 1 (Bx 1 C) 3x 2 1
5
(3x 2 1)(x 2 1 1)
11x 2 2 14x 1 9 5 (A 1 3B)x 2 1 (3C 2 B)x
1 (A 2 C)
A 2 C 5 9, 3C 2 B 5 214, A 1 3B 5 11
B 5 3C 1 14, A 5 C 1 9
A 1 3(3C 1 14) 5 11, A 1 9C 5 231
(C 1 9) 1 9C 5 231
10C 5 240, C 5 24
B 5 3(24) 1 14 5 2, A 5 (24) 1 9 5 5
A 5 5, B 5 2, C 5 24
26.> a. The vector
EF 5 (21, 24, 26) 2 (4, 0, 3)
5 (25, 24, 23)
This is a direction vector for the line containing the
segment EF. The point E(21, 24, 26) is on this
line, so the vector equation of this line is
>
r 5 (21, 24, 26) 1 t(25, 24, 23), tPR.
b. Based on the equation of the line found in part a.,
a general point on this line is of the form
J(21 2 5t, 24 2 4t, 26 2 3t), tPR.
For> this general point, the vector
JD 5 (3, 0, 7) 2 (21 2 5t, 24 24t, 26 2 3t)
5 (4 1 5t, 4 1 4t, 13 1 3t)
This vector will be perpendicular to the direction
vector for the line found in part a. at the point J we
seek. This means that
0 5 (4 1 5t, 4 1 4t, 13 1 3t) ? (25, 24, 23)
5 25(4 1 5t) 2 4(4 1 4t) 2 3(13 1 3t)
5 275 2 50t
3
t52
2
Substituting this value of t into the equation for the
general point on the line in part a.,
J(21 2 5t, 24 2 4t, 26 2 3t)
3
3
3
5 Ja21 2 5a2 b, 24 2 4a2 b, 26 2 3a2 bb
2
2
2
13
3
5 a , 2, 2 b
2
2
These are the coordinates for the point J we wanted.
c. Using the coordinates for J found in part b.,
>
13
3
JD 5 (3, 0, 7) 2 a , 2, 2 b
2
2
7
17
5 a2 , 22, b
2
2
25.
Calculus and Vectors Solutions Manual
This vector forms the height of ^ DEF, and the
length of this vector is
>
7
17
@ JD @ 5 ` a2 , 22, b `
2
2
2
7
17 2
5 a2 b 1 (22)2 1 a b
Å
2
2
177
Å 72
8 9.41
The length of the base of ^ DEF is
>
@ EF @ 5 0 (25, 24, 23) 0
5
5 "(25)2 1 (24)2 1 (23)2
5 "50
8 7.07
So the area of ^ DEF equals
177
5
1
( !50) a
b 5 !177
2
Å 2
2
8 33.26 units2
27. 3x 2 2z 1 1 5 0
4x 1 3y 1 7 5 0
(5, 25, 5)
>
>
n1 3 n2 5 (3, 0, 22) 3 (4, 3, 0) 5 (6, 28, 9)
6x 2 8y 1 9z 1 D 5 0
D 5 2115
6x 2 8y 1 9z 2 115 5 0
Chapter 9 Test, p. 556
>
1. a. r 1 5 (4, 2, 6) 1 s(1, 3, 11), sPR,
>
r 2 5 (5, 21, 4) 1 t(2, 0, 9), tPR
L1: x 5 4 1 s, y 5 2 1 3s, z 5 6 1 11s
L2: x 5 5 1 2t, y 5 21, z 5 4 1 9t
y 5 21 5 2 1 3s
s 5 21
L1: x 5 4 1 (21), y 5 2 1 3(21),
z 5 6 1 11(21)
x 5 3, y 5 21, z 5 25
(3, 21, 25)
b.
x2y1z1150
3 2 (21) 1 (25) 1 1 5 0
311251150
050
2. Use the distance equation.
0 Ax0 1 By0 1 Cz0 1 D 0
d5
"A 2 1 B 2 1 C 2
a. A(3, 2, 3)
8x 2 8y 1 4z 2 7 5 0
9-33
d5
5
0 8x0 2 8y0 1 4z0 2 7 0
"(8) 1 (28) 1 (4)
0 8(3) 2 8(2) 1 4(3) 2 7 0
2
2
Equation 2 1 (2 3 equation
4x 1 y 5 21
Equation 2 1 (8 3 equation
2
"(8)2 1 (28)2 1 (4)2
13
5
or 1.08
12
b. First, find any point on one of the planes, then use
the other plane equation with the distance formula.
2x 2 y 1 2z 2 16 5 0
2x 2 y 1 2z 1 24 5 0
2(8) 2 (0) 1 2(0) 2 16 5 0
A(8, 0, 0)
0 2x0 2 1y0 1 2z0 1 24 0
d5
"(2)2 1 (21)2 1 (2)2
0 2(8) 2 1(0) 1 2(0) 1 24 0
5
"(2)2 1 (21)2 1 (2)2
40
5
or 13.33
3
3. a. L1: 2x 1 3y 2 z 5 3
L2: 2x 1 y 1 z 5 1
L1 1 2L2: 5y 1 z 5 5
z 5 t,
5y 1 (t) 5 5
t
y512
5
2x 1 y 1 z 5 1
t
2x 1 a1 2 b 1 (t) 5 1
5
4t
x5
5
t
4t
x 5 , y 5 1 2 , z 5 t, tPR
5
5
b. To determine the point of intersection with the
xz-plane, set the above y parametric equation equal
to 0 and solve for the t. This t corresponds to the
point of intersection.
t
4t
x 5 , y 5 1 2 , z 5 t, tPR
5
5
t
0512
5
t55
(5)
4(5)
x5
,y512
, z 5 (5), tPR
5
5
(4, 0, 5)
4. a. 1
x 2 y 1 z 5 10
2 2x 1 3y 2 2z 5 221
2
1
1
1
3
x1 y1 z52
2
5
4
2
9-34
6x 1
1)
5
3)
5
31
y 5 225
5
31
(4x 1 y 5 21)
5
31
1 a6x 1 y 5 225b
5
218.8x 5 218.8
x51
4(1) 1 y 5 21
y 5 25
(1) 2 (25) 1 z 5 10
z54
(1, 25, 4)
b. The three planes intersect at this point.
1
5. a.
x 2 y 1 z 5 21
2 2x 1 2y 2 z 5 0
3 x 2 5y 1 4z 5 23
Equation 2 1 (2 3 equation 1 ) 5
4x 1 z 5 22
4x 1 z 5 22
z5t
4x 1 (t) 5 22
1
t
x52 2
2
4
x 2 y 1 z 5 21
1
t
a2 2 b 2 y 1 (t) 5 21
2
4
3t
1
y5 1
4
2
3t
1
t
1
x 5 2 2 , y 5 1 , z 5 t, tPR
2
4
4
2
b. The three planes intersect at this line.
6. a. L1: x 1 y 1 z 5 0
L2: x 1 2y 1 2z 5 1
L3: 2x 2 y 1 mz 5 n
L2 1 2L3: 5x 1 0y 1 (2m 1 2)z 5 2n 1 1
L1 1 L3: 3x 1 0y 1 (m 1 1)z 5 n
5
(3x 1 0y 1 (m 1 1)z 5 n)
3
5
5
5 5x 1 0y 1 (m 1 1)z 5 n
3
3
Then set the two new equations to each other and
solve for a m and n value that would give equivalent
equations.
2
Chapter 9: Relationships Between Points, Lines, and Planes
5
n
3
2n 1 1
5
2m 1 2 5 (m 1 1)
3
m 5 21
5
n 5 2n 1 1
3
n 5 23
b. L1: x 1 y 1 z 5 0
L2: x 1 2y 1 2z 5 1
L3: 2x 2 y 2 z 5 23
L1 1 L2: 3x 5 23, x 5 21
(21) 1 y 1 z 5 0
z5t
(21) 1 y 1 (t) 5 0
y512t
x 5 21, y 5 1 2 t, z 5 t, tPR
7. First find the parametric equations of each line.
Then set these equations equal to each other to find
a set of new equations. Use the dot product to
determine another set of equations that you will
solve for t and s. Find the corresponding points to
these values and the distance between them, which
is the distance between the two lines.
>
L1: r 5 (21, 23, 0) 1 s(1, 1, 1), sPR
>
L2: r 5 (25, 5, 28) 1 t(1, 2, 5), tPR
L1: x 5 21 1 s, y 5 23 1 s, z 5 s
L2: x 5 25 1 t, y 5 5 1 2t, z 5 28 1 5t
>
UV 5 3 (21 1 s) 2 (25 1 t), (23 1 s)
2 (5 1 2t), s 2 (28 1 5t)4
>
UV 5 (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t)
>
m1 ? UV 5 0
(1, 1, 1) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
(1, 2, 5) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
L4: 4 1 3s 2 8t 5 0,
L5: 28 1 8s 2 30t 5 0
8 3 L1 1 (23) 3 L2 yields
32 1 24s 2 64t 2 84 2 24s 1 90t 5 0, so t 5 2.
Then s 5 4. The points corresponding to these
values of s and t are (21, 3, 0) 1 4(1, 1, 1)
5 (3, 1, 4) and (25, 5, 28) 1 2(1, 2, 5)
5 (23, 9, 2).
5
5x 1 0y 1 (m 1 1)z 5
3
5x 1 0y 1 (2m 1 2)z 5
d 5 "(3 2 (23))2 1 (1 2 9)2 1 (4 2 2)2
5 "(6)2 1 (28)2 1 (2)2
5 "36 1 64 1 4
5 "104 or 10.20
Calculus and Vectors Solutions Manual
Cumulative Review of Vectors,
pp. 557–560
1. a. The angle, u, between the two
vectors is found
>
>
a?b
from the equation cos (u) 5 > > .
@a @ @b@
> >
a ? b 5 (2, 21, 22) ? (3, 24, 12)
5 2(3) 2 1(24) 2 2(12)
5 214
>
0 a 0 5 "22 1 (21)2 1 (22)2
53
0 b 0 5 "32 1 (24)2 1 122
5 13
So u 5 cos21 ( 3 214
3 13 )
8 111.0°
>
>
>
b. The scalar projection of a on b is equal to
>
0 a 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 13
>
>
>
and 0 a 0 5 3, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 3 5 2 13 . The vector projection of a on b
is equal to the scalar projection multiplied by the unit
>
vector in the direction of b. So the vector projection
1
52 56
, , 2 168
is 2 14
13 3 13 (3, 24, 12) 5 (2
169 ).
> 169 169
>
c. > The scalar projection of b on a is equal to
0 b 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 > 13
>
>
and 0 b 0 5 13, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 13 5 2 3 . The vector projection of b on a
is equal to the scalar projection multiplied by the
>
unit vector in the direction of a . So the vector
projection is 2 143 3 13 (2, 21, 22) 5 (2 289, 149, 289 ).
2. a. Since the normal of the first plane is (4, 2, 6)
and the normal of the second is (1, 21, 1), which
are not scalar multiples of each other, there is a line
of intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The first equation minus four times the second
equation yields 0x 1 6y 1 2z 1 6 5 0. We may
divide by two to simplify, so 3y 1 z 1 3 5 0. If we
let y 5 t, then 3t 1 z 1 3 5 0, or z 5 23 2 3t.
Substituting these into the second equation yields
x 2 (t) 1 (23 2 3t) 2 5 5 0 or x 5 8 1 4t.
So the equation of the line in parametric form is
x 5 8 1 4t, y 5 t, z 5 23 2 3t, tPR.
9-35
5
n
3
2n 1 1
5
2m 1 2 5 (m 1 1)
3
m 5 21
5
n 5 2n 1 1
3
n 5 23
b. L1: x 1 y 1 z 5 0
L2: x 1 2y 1 2z 5 1
L3: 2x 2 y 2 z 5 23
L1 1 L2: 3x 5 23, x 5 21
(21) 1 y 1 z 5 0
z5t
(21) 1 y 1 (t) 5 0
y512t
x 5 21, y 5 1 2 t, z 5 t, tPR
7. First find the parametric equations of each line.
Then set these equations equal to each other to find
a set of new equations. Use the dot product to
determine another set of equations that you will
solve for t and s. Find the corresponding points to
these values and the distance between them, which
is the distance between the two lines.
>
L1: r 5 (21, 23, 0) 1 s(1, 1, 1), sPR
>
L2: r 5 (25, 5, 28) 1 t(1, 2, 5), tPR
L1: x 5 21 1 s, y 5 23 1 s, z 5 s
L2: x 5 25 1 t, y 5 5 1 2t, z 5 28 1 5t
>
UV 5 3 (21 1 s) 2 (25 1 t), (23 1 s)
2 (5 1 2t), s 2 (28 1 5t)4
>
UV 5 (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t)
>
m1 ? UV 5 0
(1, 1, 1) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
(1, 2, 5) ? (4 1 s 2 t, 28 1 s 2 2t, s 1 8 2 5t) 5 0
L4: 4 1 3s 2 8t 5 0,
L5: 28 1 8s 2 30t 5 0
8 3 L1 1 (23) 3 L2 yields
32 1 24s 2 64t 2 84 2 24s 1 90t 5 0, so t 5 2.
Then s 5 4. The points corresponding to these
values of s and t are (21, 3, 0) 1 4(1, 1, 1)
5 (3, 1, 4) and (25, 5, 28) 1 2(1, 2, 5)
5 (23, 9, 2).
5
5x 1 0y 1 (m 1 1)z 5
3
5x 1 0y 1 (2m 1 2)z 5
d 5 "(3 2 (23))2 1 (1 2 9)2 1 (4 2 2)2
5 "(6)2 1 (28)2 1 (2)2
5 "36 1 64 1 4
5 "104 or 10.20
Calculus and Vectors Solutions Manual
Cumulative Review of Vectors,
pp. 557–560
1. a. The angle, u, between the two
vectors is found
>
>
a?b
from the equation cos (u) 5 > > .
@a @ @b@
> >
a ? b 5 (2, 21, 22) ? (3, 24, 12)
5 2(3) 2 1(24) 2 2(12)
5 214
>
0 a 0 5 "22 1 (21)2 1 (22)2
53
0 b 0 5 "32 1 (24)2 1 122
5 13
So u 5 cos21 ( 3 214
3 13 )
8 111.0°
>
>
>
b. The scalar projection of a on b is equal to
>
0 a 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 13
>
>
>
and 0 a 0 5 3, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 3 5 2 13 . The vector projection of a on b
is equal to the scalar projection multiplied by the unit
>
vector in the direction of b. So the vector projection
1
52 56
, , 2 168
is 2 14
13 3 13 (3, 24, 12) 5 (2
169 ).
> 169 169
>
c. > The scalar projection of b on a is equal to
0 b 0 cos (u), where u is the angle between the two
vectors. So from the above work, cos (u) 5 3 214
3 > 13
>
>
and 0 b 0 5 13, so the scalar projection of a on b is
>
>
214
14
3 3 13 3 13 5 2 3 . The vector projection of b on a
is equal to the scalar projection multiplied by the
>
unit vector in the direction of a . So the vector
projection is 2 143 3 13 (2, 21, 22) 5 (2 289, 149, 289 ).
2. a. Since the normal of the first plane is (4, 2, 6)
and the normal of the second is (1, 21, 1), which
are not scalar multiples of each other, there is a line
of intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The first equation minus four times the second
equation yields 0x 1 6y 1 2z 1 6 5 0. We may
divide by two to simplify, so 3y 1 z 1 3 5 0. If we
let y 5 t, then 3t 1 z 1 3 5 0, or z 5 23 2 3t.
Substituting these into the second equation yields
x 2 (t) 1 (23 2 3t) 2 5 5 0 or x 5 8 1 4t.
So the equation of the line in parametric form is
x 5 8 1 4t, y 5 t, z 5 23 2 3t, tPR.
9-35
To check that this is correct, we substitute in the
solution to both initial equations
4x 1 2y 1 6z 2 14 5 4(8 1 4t) 1 2(t)
1 6(23 2 3t) 2 14
50
and x 2 y 1 z 2 5
5 (8 1 4t) 2 (t) 1 (23 2 3t) 2 5
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The angle between two planes is the same as the
angle between their corresponding normal vectors.
0 (4, 2, 6)0 5 "42 1 22 1 62
5 !56
0 (1, 21, 1) 0 5 "12 1 12 1 12
5 !3
(4, 2, 6) ? (1, 21, 1) 5 8, so the angle between the
planes is cos21 Q !38!56 R 8 51.9°.
> >
>
x?y
3. a. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since x
>
>
>
and y are unit vectors, 0 x 0 5 0 y 0 5 1, and moreover
> >
> >
1
x?y
1
cos (60°) 5 . So x ? y 5
5 .
2
131
2
b. Scalar multiples can be brought out to the front
>
>
> >
of dot products. Hence 2x ? 3y 5 (2)(3)(x ? y ),
>
>
and so by part a., 2x ? 3y 5 2 3 3 3 12 5 3.
c. The dot product is distributive,
>
>
>
>
so (2x 2 y ) ? (x 1 3y )
>
>
>
>
>
>
5 2x ? (x 1 3y ) 2 y ? (x 1 3y )
> >
>
>
> >
>
>
5 2x ? x 1 2x ? 3y 2 y ? x 2 y ? 3y
> >
>
>
> >
> >
5 2x ? x 1 2x ? 3y 2 x ? y 2 3y ? y
>
>
> >
> >
Since x and y are unit vectors, x ? x 5 y ? y 5 1,
and so by using the values found in part a. and b.,
>
>
>
>
(2x 2 y ) ? (x 1 3y ) 5 2(1) 1 (3) 2 A 12 B 2 3(1)
3
5
2 >
>
>
>
>
>
>
>
4. a. 2(i 2 2j 1 3k ) 2 4(2i 1 4j 1 5k ) 2 (i 2 j )
>
>
>
>
>
>
>
>
5 2i 2 4j 1 6k 2 8i 2 16j 2 20k 2 i 1 j
>
>
>
5 27i 2 19j 2 14k
>
>
>
>
>
>
>
>
b. 22(3i> 2 4j> 2 5k>) ? (2i> 1 3k> ) 1 2i
?
(3j
2
2k
)
>
5 22(3i 2 4j 2 5k ) ? (2i 1 0j 1 3k )
>
>
>
>
>
>
1 2(i 1 0j 1 0k ) ? (0i 1 3j 2 2k )
5 22(3(2) 2 4(0) 2 5(3)) 1 2(1(0)
1 0(3) 1 0(22))
5 22(29) 1 2(0)
5 18
9-36
5. The direction vectors for the positive x-axis,
y-axis, and z-axis are (1, 0, 0), (0, 1, 0), and (0, 0, 1),
respectively.
0 (4, 22, 23) 0 5 "42 1 (22)2 1 (23)2
5 !29,
and 0 (1, 0, 0) 0 5 0 (0, 1, 0) 0
5 0 (0, 0, 1) 0
5 !1
5 1.
(4, 22, 23) ? (1, 0, 0) 5 4, so the angle the vector
4
makes with the x-axis is cos21 Q 1 !29
R 8 42.0°.
(4, 22, 23) ? (0, 1, 0) 5 22, so the angle the vector
makes with the y-axis is cos21 Q 1 22
!29 R 5 111.8°.
(4,22,23) ? (0, 0, 1) 5 23, hence the angle the
vector makes with the z-axis is cos21 Q 1 23
!29 R 8 123.9°.
>
>
6. a. a 3 b 5 (1, 22, 3) 3 (21, 1, 2)
5 (22(2) 2 3(1), 3(21) 2 1(2),
1(1) 2 (22)(21))
5 (27, 25, 21)
b. By the> scalar law for> vector multiplication,
>
>
2a 3 3b 5 2(3)(a 3
> b)
>
5 6(a 3 b )
5 6(27, 25, 21) 5 (242, 230, 26)
>
c.> The area of a parallelogram determined by a and
b is equal
> to the magnitude of the cross product of
>
a and b.
A 5 area of> parallelogram
>
5 0a 3 b0
5 0 (27, 25, 21) 0
5 "(27)2 1 (25)2 1 (21)2
8 >8.66 square units >
>
>
d. (b 3 a ) 5 2 (a 3 b )
5 2 (27, 25, 21)
5 (7, 5, 1)
>
>
>
So c ? (b 3 a ) 5 (3, 24, 21) ? (7, 5, 1)
5 3(7) 2 4(5) 2 1(1)
50
>
>
7. A unit vector perpendicular to both a and b can
be determined
> from
> any vector perpendicular to
>
>
both a and b>. a 3 b is a vector perpendicular to
>
both a and
b.
>
>
a 3 b 5 (1, 21, 1) 3 (2, 22, 3)
5 (21(3) 2 1(22), 1(2) 2 1(3),
1(22) 2 (21)(2))
5 (21, 21, 0)
Chapter 9: Relationships Between Points, Lines, and Planes
>
>
0 a 3 b 0 5 0 (21, 21, 0) 0
5 "(21)2 1 (21)2 1 02
5 !2
1
1
1
So !2
(21, 21, 0) 5 Q 2 !2
, 2 !2
, 0 R is an unit vector
>
>
1
1
, !2
, 0 R is another.
perpendicular to both a and b. Q !2
8. a. Answers may vary. For example:
>
A direction
vector for the line is AB.
>
AB 5 (1, 2, 3) 2 (2, 23, 1)
5 (21, 5, 2)
Since A(2, 23, 1) is a point on the line,
>
r 5 (2, 23, 1) 1 t(21, 5, 2), tPR, is a vector
equation for a line and the corresponding parametric
equation is x 5 2 2 t, y 5 23 1 5t, z 5 1 1 2t,
tPR.
b. If the x-coordinate of a point on the line is 4, then
2 2 t 5 4, or t 5 22. At t 5 22, the point on the
line is (2, 23, 1) 2 2(21, 5, 2) 5 (4, 213, 23).
Hence C(4, 213, 23) is a point on the line.
9. The direction vector of the first line is (21, 5, 2),
while the direction vector for the second line is
(1, 25, 22) 5 2 (21, 5, 2). So the direction vectors
for the line are collinear. Hence the lines are parallel.
The lines coincide if and only if for any point on
the first line and any point on the second line, the
vector connecting the two points is a multiple of the
direction vector for the lines.
(2, 0, 9) is a point on the first line and (3, 25, 10) is
a point on the second line.
(2, 0, 9) 2 (3, 25, 10) 5 (21, 5, 21) 2 k(21, 5, 2)
for any kPR. Hence the lines are parallel and distinct.
10. The direction vector for the parallel line is
(0, 1, 1). Since parallel lines have collinear direction
vectors, (0, 1, 1) can be used as a direction vector
for the line. Since (0, 0, 4) is a point on the line,
>
r 5 (0, 0, 4) 1 t(0, 1, 1), tPR, is a vector equation
for a line and the corresponding parametric equation
is x 5 0, y 5 t, z 5 4 1 t, tPR.
11. The line is parallel to the plane if and only if the
direction vector for the line is perpendicular to the
normal vector for the plane. The normal vector for
the plane is (2, 3, c). The direction vector for the
line is (2, 3, 1). The vectors are perpendicular if and
only if the dot product between the two is zero.
(2, 3, c) ? (2, 3, 1) 5 2(2) 1 3(3) 1 c(1)
5 13 1 c
So if c 5 213, then the dot product of normal
vector and the direction vector is zero. Hence for
c 5 213, the line and plane are parallel.
Calculus and Vectors Solutions Manual
12. First put the line in its corresponding parametric
form. (3, 1, 5) is a direction vector and (2, 25, 3) is
the origin point, so a parametric equation for the
line is x 5 2 1 3s, y 5 25 1 s, z 5 3 1 5s, sPR.
If we substitute these coordinates into the equation
of the plane, we may find the s value where the line
intersects the plane.
5x 1 y 2 2z 1 2
5 5(2 1 3s) 1 (25 1 s) 2 2(3 1 5s) 1 2
5 10 1 15s 1 2 5 1 s 2 6 2 10s 1 2
5 1 1 6s
So if 5x 1 y 2 2z 1 2 5 0, then 1 1 6s 5 0 or
s 5 2 16. At s 5 2 16, the point on the line is ( 32, 2 316, 136) .
To check that this point is also on the plane, we
substitute the x, y, z values into the plane equation
and check that it equals zero.
3
31
13
5x 1 y 2 2z 1 2 5 5a b 1 a2 b 2 2a b 1 2
2
6
6
50
Hence ( 32, 2 316, 136) is the point of intersection between
the line and the plane.
13. a.
z
(0, 0, 3)
(0, 3, 0)
x
y
(6, 0, 0)
Two direction vectors are:
(0, 3, 0) 2 (0, 0, 3) 5 (0, 3, 23)
and
(6, 0, 0) 2 (0, 0, 3) 5 (6, 0, 23).
9-37
b.
z
(–3, –2, 2)
(0, 0, 0)
y
(3, 2, 1)
x
Two direction vectors are:
(23, 22, 2) 2 (0, 0, 0) 5 (23, 22, 2)
and
(3, 2, 1) 2 (0, 0, 0) 5 (3, 2, 1).
c.
z
(0, 3, 6)
(0, 0, 0)
x
(1, 1, –1)
y
Two direction vectors are:
(0, 3, 6) 2 (0, 0, 0) 5 (0, 3, 6)
and
(1, 1, 21) 2 (0, 0, 0) 5 (1, 1, 21).
14. The plane is the right bisector joining
P(1, 22, 4) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (2, 23, 24). So the line connecting the
two points is (1, 22, 4) 1 t(2, 23, 24), tPR, or in
9-38
corresponding parametric form is x 5 1 1 2t,
y 5 22 2 3t, z 5 4 2 4t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we substitute the parametric equation into the plane
equation and solve for t.
2x 2 3y 2 4z 1 66
5 2(1 1 2t) 2 3(22 2 3t) 2 4(4 2 4t) 1 66
5 2 1 4t 1 6 1 9t 2 16 1 16t 1 66
5 58 1 29t
So if 2x 2 3y 2 4z 1 66 5 0, then 58 1 29t 5 0,
or t 5 22.
So the point of intersection is occurs at t 5 22, since
the origin point is P and the intersection occurs at the
midpoint of the line connecting P and its image, the
image point occurs at t 5 2 3 (22) 5 24.
So the image point is at x 5 1 1 2(24) 5 27,
y 5 22 2 3(24) 5 10, z 5 4 2 4(24) 5 20.
So the image point is (27, 10, 20).
15. Let (a, b, c) be the direction vector for this line.
>
So a line equation is r 5 (1, 0, 2) 1 t(a, b, c), tPR.
Since (1, 0, 2) is not on the other line, we may
choose a, b, and c such that the intersection occurs
at t 5 1. Since the line is supposed to intersect the
given line at a right angle, the direction vectors
should be perpendicular. The direction vectors are
perpendicular if and only if their dot product is zero.
The direction vector for the given line is (1, 1, 2).
(a, b, c) ? (1, 1, 2) 5 a 1 b 1 2c 5 0, so
b 5 2a 2 2c.
Also (1, 0, 2) 1 (a, b, c) 5 (1 1 a, b, 2 1 c) is the
point of intersection.
By substituting for b,
(1 1 a, b, 2 1 c) 5 (1 1 a, 2a 2 2c, 2 1 c).
So for some s value,
x 5 22 1 s 5 1 1 a
y 5 3 1 s 5 2a 2 2c
z 5 4 1 2s 5 2 1 c
Subtracting the first equation from the second yields
the equation, 5 1 0s 5 22a 2 2c 2 1.
Simplifying this gives 6 5 22a 2 2c or just
a 1 c 5 23.
Subtracting twice the first equation from the third
yields the equation, 8 5 22a 1 c.
So a 1 c 5 23 and 22a 1 c 5 8, which is two
equations with two unknowns. Twice the first plus
the second equations gives 0a 1 3c 5 2 or c 5 23.
Solving back for a gives 2 113 and since b 5 2a 2 2c,
b 5 73. Since a 1 b 1 2c 5 0, the direction vectors,
Chapter 9: Relationships Between Points, Lines, and Planes
(1, 1, 2) and (a, b, c) are perpendicular. A direction
vector for the line is (211, 7, 2).
We need to check that
7 8
(1, 0, 2) 1 (a, b, c) 5 ( 28
3 , 3 , 3 ) is a point on the
given line.
x 5 22 1 s 5 2 83, at s 5 2 23. The point on the given
7 8
line at s 5 2 23 is Q 28
3 , 3 , 3 R . Hence
>
q 5 (1, 0, 2) 1 t(211, 7, 2), tPR, is a line that
intersects the given line at a right angle.
16. a. The Cartesian equation is found by taking
> the
cross
> product of the two direction vectors, AB and
AC.
>
AB 5 (22, 0, 0) 2 (1, 2, 3)
> 5 (23, 22, 23)
AC 5 (1, 4, 0) 2 (1, 2, 3) 5 (0, 2, 23)
>
>
AB 3 AC 5 (22(23) 2 (23)(2),
23(0) 2 (23)(23),
23(2) 2 (22)(0))
5 (12, 29, 26)
So 5 (12, 29, 26) is a normal vector for the
plane, so the plane has the form
12x 2 9y 2 6z 1 D 5 0, for some constant D. To
find D, we know that A(1, 2, 3) is a point on the
plane, so 12(1) 2 9(2) 2 6(3) 1 D 5 0. So
224 1 D 5 0, or D 5 24. So the Cartesian
equation for the plane is 12x 2 9y 2 6z 1 24 5 0.
b. Substitute into the formula to determine distance
between a point and a plane. So the distance, d, of
(0, 0, 0) to the plane 12x 2 9y 2 6z 1 24 5 0 is
equal to
@ 12 (0) 2 9 (0) 2 6 (0) 1 24 @
"122 1 (29)2 1 (26)2
24
!261 8 1.49.
c. Since the plane is perpendicular to the z-axis, a
direction vector for the z-axis acts as a normal vector
for the plane. Hence (0, 0, 1) is a normal vector for
the plane. So the plane has the form z 1 D 5 0, for
some constant D. To find D, we know that (3, 21, 3)
is a point on the plane, so
0(3) 1 0(21) 1 (3) 1 D 5 0. So 3 1 D 5 0, or
D 5 23. So the Cartesian equation for the plane is
z 2 3 5 0.
d. The Cartesian equation can be found by taking
the cross product of the two direction vectors for
the plane. Since (3, 1, 22) and (1, 3, 21) are two
points on the plane
(3, 1, 22) 2 (1, 3, 21) 5 (2, 22, 21) is a
direction vector for the plane. Since the plane is
parallel to the y-axis, (0, 1, 0) is also a direction
vector for the plane.
(2, 22, 21) 3 (0, 1, 0) 5 (22(0) 2
(21)(1), (21)(0)2 (2)(0), 2(1) 2 (22)(0))
5 (1, 0, 2)
So (1, 0, 2) is a normal vector for the plane, so the
plane has the form x 1 0y 1 2z 1 D 5 0, for some
constant D. To find D, we know that (3, 1, 22) is a
point on the plane, so
(3) 1 0(1) 1 2(22) 1 D 5 0. So 21 1 D 5 0,
or D 5 1. So the Cartesian equation for the plane is
x 1 2z 1 1 5 0.
18.
E
100 km/h
45°
F
.
So d 5
17. a. (3, 25, 4) is a normal vector for the plane, so
the plane has the form 3x 2 5y 1 4z 1 D 5 0, for
some constant D. To find D, we know that
A(21, 2, 5) is a point on the plane, so
3(21) 2 5(2) 1 4(5) 1 D 5 0. So 7 1 D 5 0,
or D 5 27. So the Cartesian equation for the plane
is 3x 2 5y 1 4z 2 7 5 0.
b. Since the plane is perpendicular to the line
connecting (2, 1, 8) and (1, 2, 24), a direction
vector for the line acts as a normal vector for the
plane. So (2, 1, 8) 2 (1, 2, 24) 5 (1, 21, 12) is a
normal vector for the plane. So the plane has the
form x 2 y 1 12z 1 D 5 0, for some constant D.
To find D, we know that K(4, 1, 2) is a point on the
plane, so (4) 2 (1) 1 12(2) 1 D 5 0. So
27 1 D 5 0, or D 5 227. So the Cartesian
equation for the plane is x 2 y 1 12z 2 27 5 0.
Calculus and Vectors Solutions Manual
400 km/h
45°
100 km/h
Position Diagram
R
400 km/h
D
Vector Diagram
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 4002 1 1002 2 2(400)(100) cos (45°)
8 336.80 km> h.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin 45°
sin /EDF
8
.
336.80
100
9-39
sin 45°
3 100.
336.80
sin /EDF 8 0.2100.
Thus /EDF 8 12.1°, so the resultant velocity is
336.80 km> h, N 12.1° W.
19. a. The simplest way is to find the parametric
equation, then find the corresponding vector equation.
If we substitute x 5 s and y 5 t and solve for z, we
obtain 3s 2 2t 1 z 2 6 5 0 or z 5 6 2 3s 1 2t.
This yields the parametric equations x 5 s, y 5 t,
and z 5 6 2 3s 1 2t. So the corresponding vector
>
equation is r 5 (0, 0, 6) 1 s(1, 0, 23) 1 t(0, 1, 2),
s, tPR. To check that this is correct, find the
Cartesian equation corresponding to the above
vector equation and see if it is equivalent to the
Cartesian equation given in the problem. A normal
vector to this plane is the cross product of the two
directional
vectors.
>
n 5 (1, 0, 23) 3 (0, 1, 2) 5 (0(2) 2 (23)(1),
23(0) 2 1(2), 1(1) 2 0(0))
5 (3, 22, 1)
So (3, 22, 1) is a normal vector for the plane, so the
plane has the form 3x 2 2y 1 z 1 D 5 0, for some
constant D. To find D, we know that (0, 0, 6) is a point
on the plane, so 3(0) 2 2(0) 1 (6) 1 D 5 0.
So 6 1 D 5 0, or D 5 26. So the Cartesian equation
for the plane is 3x 2 2y 1 z 2 6 5 0. Since this is
the same as the initial Cartesian equation, the vector
equation for the plane is correct.
b.
z
sin /EDF 8
(0, 0, 6)
(0, –3, 0)
vector for the plane. The direction vector of the line
is (2, 21, 2) and the normal vector for the plane
is (1, 2, 1).
0 (2, 21, 2) 0 5 "22 1 (21)2 1 22
5 !9
5 3.
0 (1, 2, 1) 0 5 "12 1 22 1 12
5 !6
(2, 21, 2) ? (1, 2, 1) 5 2(1) 2 1(2) 1 2(1) 5 2
So the angle between the normal vector and the
2
direction vector is cos21 Q 3 !6
R 8 74.21°. So
u 8 90° 2 74.21° 5 15.79°.
To the nearest degree, u 5 16°.
b. The two planes are perpendicular if and only if
their normal vectors are also perpendicular.
A normal vector for the first plane is (2, 23, 1) and
a normal vector for the second plane is
(4, 23, 217). The two vectors are perpendicular if
and only if their dot product is zero.
(2, 23, 1) ? (4, 23, 217) 5 2(4) 2 3(23)
1 1(217)
5 0.
Hence the normal vectors are perpendicular. Thus
the planes are perpendicular.
c. The two planes are parallel if and only if their
normal vectors are also parallel. A normal vector for
the first plane is (2, 23, 2) and a normal vector for
the second plane is (2, 23, 2). Since both normal
vectors are the same, the planes are parallel. Since
2(0) 2 3(21) 1 2(0) 2 3 5 0, the point
(0, 21, 0) is on the second plane. Yet since
2(0) 2 3(21) 1 2(0) 2 1 5 2 2 0, (0, 21, 0) is
not on the first plane. Thus the two planes are
parallel but not coincident.
21.
25 N
y
x
60°
(2, 0, 0)
40 N
Position diagram
E
40 N
F
120°
20. a. The angle, u, between the plane and the line
is the complementary angle of the angle between
the direction vector of the line and the normal
9-40
25 N
60°
D
R
120°
40 N
Vector diagram
Chapter 9: Relationships Between Points, Lines, and Planes
From
the triangle DEF and the cosine law, we have
>
0 R 0 2 5 402 1 252 2 2(40)(25) cos (120°)
8 56.79 N.
To find the direction of the vector, the sine law is
applied.
sin /DEF
sin /EDF
>
5
100
0R0
sin 120°
sin /EDF
8
.
56.79
40
sin 120°
sin /EDF 8
3 40.
56.79
sin /EDF 8 0.610.
Thus /EDF 8 37.6°, so the resultant force is
approximately 56.79 N, 37.6° from the 25 N force
towards the 40 N force. The equilibrant force has
the same magnitude as the resultant, but it is in
the opposite direction. So the equilibrant is
approximately 56.79 N, 180° 2 37.6° 5 142.4°
from the 25 N force away from the 40 N force.
22.
a
b
–b
–b
a
a –b
b.
1
2
2a
1
2
b
2a +
1
2
b
b
2a
>
23. a. The unit vector in the same direction of a is
>
>
simply a divided by the magnitude of a .
>
0 a 0 5 "62 1 22 1 (23)2
5 "49
57
>
So the unit vector in the same direction of a is
1 >
> a 5 17 (6, 2, 23) 5 ( 67, 27, 2 37 ).
0a0
>
b. The unit vector in the opposite direction of a is
simply the negative of the unit vector found in part
a. So the vector is 2 A 67, 27, 2 37 B 5 A2 67, 2 27, 37 B.
24. a. Since OBCD is a parallelogram, the point
C
>
occurs at (21, 7) 1 (9, 2) 5 (8, 9). So> OC is one
vector
equivalent to a diagonal and BD is the other.
>
OC> 5 (8, 9) 2 (0, 0) 5 (8, 9)
BD 5 (9, 2) 2 (1, 7) 5 (10, 25)
Calculus and Vectors Solutions Manual
0 (8, 9) 0 5 "82 1 92
b.
5 "145
0 (10, 25) 0 5 "102 1 (25)2
5 "125
(8, 9) ? (10, 25) 5 8(10) 1 9(25)
5 235
So the angle between these diagonals is
235
cos21 A !145
!125 B 8 74.9°.
>
>
c. OB 5 (21, 7) and OD 5 (9, 2)
0 (21, 7) 0 5 "(21)2 1 72.
5 "50
0 (9, 2) 0 5 "92 1 22
5 "85
(21, 7) ? (9, 2) 5 2 (9) 1 7(2)
55
So the angle between these diagonals is
cos21 A !505!85 B 8 85.6°.
25. a. First step is to use the first equation to
remove x from the second and third.
1 x2y1z52
2 2x 1 y 1 2z 5 1
3 x 2 y 1 4z 5 5
So we have
4 0x 1 0y 1 3z 5 3, 1 1 2
5 0x 1 0y 1 3z 5 3, 21 3 1 1 3
Hence 3z 5 3, or z 5 1. Since both equations are
the same, this implies that there are infinitely many
solutions. Let x 5 t, then by substituting into the
equation 2, we obtain
2t 1 y 1 2(1) 5 1, or y 5 21 1 t.
Hence the solution to these equations is x 5 t,
y 5 21 1 t, z 5 1, tPR.
b. First step is to use the first equation to remove x
from the second and third.
1 22x 2 3y 1 z 5 211
2 x 1 2y 1 z 5 2
3 2x 2 y 1 3z 5 212
So we have
4 0x 1 1y 1 3z 5 27, 1 1 2 3 2
5 0x 2 1y 2 5z 5 13, 1 2 2 3 3
Now the fourth and fifth equations are used to
create a sixth equation where the coefficient of
y is zero.
6 0x 1 0y 2 2z 5 6, 4 1 5
So 22z 5 6 or z 5 23.
9-41
Substituting this into equation 4 yields,
y 1 3(23) 5 27 or y 5 2. Finally substitute z and
y values into equation 2 to obtain the x value.
x 1 2(2) 1 (23) 5 2 or x 5 1.
Hence the solution to these three equations is
(1, 2, 23).
c. First step is to notice that the second equation is
simply twice the first equation.
1 2x 2 y 1 z 5 21
2 4x 2 2y 1 2z 5 22
3 2x 1 y 2 z 5 5
So the solution to these equations is the same as the
solution to just the first and third equations.
Moreover since this is two equations with three
unknowns, there will be infinitely many solutions.
4 4x 1 0y 1 0z 5 4, 1 1 3
Hence 4x 5 4 or x 5 1. Let y 5 t and solve for z
using the first equation.
2(1) 2 t 1 z 5 21, so z 5 23 1 t
Hence the solution to these equations is x 5 1,
y 5 t, z 5 23 1 t, tPR.
d. First step is to notice that the second equations
is simply twice the first and the third equation is
simply 24 times the first equation.
1 x 2 y 2 3z 5 1
2 2x 2 2y 2 6z 5 2
3 24x 1 4y 1 12z 5 24
So the solution to these equations is the same as the
solution to just the first equation. So the solution to
these equations is a plane. To solve this in parametric
equation form, simply let y 5 t and z 5 s and find
the x value.
x 2 t 2 3s 5 1, or x 5 1 1 t 1 3s
So the solution to these equations is x 5 1 1 3s 1 t,
y 5 t, z 5 s, s, tPR.
26. a. Since the normal of the first equation
is (1, 21, 1) and the normal of the second is
(1, 2, 22), which are not scalar multiples of each
other, there is a line of intersection between the
planes. The next step is to use the first and second
equations to find an equation with a zero for the
coefficient of x. The second equation minus the first
equation yields 0x 1 3y 2 3z 1 3 5 0. We may
divide by three to simplify, so y 2 z 1 1 5 0. If
we let z 5 t, then y 2 t 1 1 5 0, or y 5 21 1 t.
Substituting these into the first equation yields
x 2 (21 1 t) 1 t 2 1 5 0 or x 5 0. So the
equation of the line in parametric form is x 5 0,
y 5 21 1 t, z 5 t, tPR.
9-42
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 1 5 (0) 2 (21 1 t) 1 (t) 2 1
50
and
x 1 2y 2 2z 1 2 5 (0) 1 2(21 1 t) 2 2(t) 1 2
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
b. The normal vector for the first plane is
(1, 24, 7), while the normal vector for the second
plane is (2, 28, 14) 5 2(1, 24, 7). Hence the
planes have collinear normal vectors, and so are
parallel.
The second equation is equivalent to
x 2 4y 1 7z 5 30, since we may divide the equation
by two. Since the constant on the right in the first
equation is 28, while the constant on the right in the
second equivalent equation is 30, these planes are
parallel and not coincident. So there is no intersection.
c. The normal vector for the first equation is
(1, 21, 1), while the normal vector for the second
equation is (2, 1, 1). Since the normal vectors are
not scalar multiples of each other, there is a line of
intersection between the planes.
The next step is to use the first and second equations
to find an equation with a zero for the coefficient of x.
The second equation minus twice the first equation
yields 0x 1 3y 2 z 1 0 5 0.
Solving for z yields, z 5 3y. If we let y 5 t, then
z 5 3(t) 5 3t.
Substituting these into the first equation yields
x 2 (t) 1 (3t) 2 2 5 0 or x 5 2 2 2t. So the
equation of the line in parametric form is x 5 2 2 2t,
y 5 t, z 5 3t, tPR.
To check that this is correct, we substitute in the
solution to both initial equations
x 2 y 1 z 2 2 5 (2 2 2t) 2 (t) 1 (3t) 2 2
50
and
2x 1 y 1 z 2 4 5 2(2 2 2t) 1 (t) 1 (3t) 2 4
5 0.
Hence the line given by the parametric equation
above is the line of intersection for the planes.
27. The angle, u, between the plane and the line is
the complementary angle of the angle between the
direction vector of the line and the normal vector
for the plane. The direction vector of the line is
Chapter 9: Relationships Between Points, Lines, and Planes
(1, 21, 0) and the normal vector for the plane is
(2, 0, 22).
0 (1, 21, 0) 0 5 "12 1 (21)2 1 02
5 "2
0 (2, 0, 22) 0 5 "22 1 02 1 (22)2 5 "8
(1, 21, 0) ? (2, 0, 22) 5 1(2) 2 1(0) 1 0(22) 5 2
So the angle between the normal vector and the
direction vector is cos21 A !22!8 B 5 60°. So
u 5 90 260° 5 30°.
> >
a?b
28. a. We have that cos (60°) 5 > > . Also
0a0 0b0 >
>
>
>
since a and b are unit vectors, 0 a 0 5 0 b 0 5 1 and
>
>
> >
a ? a 5 b ? b> 5 1, and moreover cos (60°) 5 12. So
>
> >
a?b
a?b5
5 12.
131
The dot >product is distributive,
so >
>
>
>
>
>
(6a 1 b ) ? (a 2 2b ) 5 6a ? (a 2 2b )
>
>
>
1 b ? (a 2 2b )
>
> >
>
5 6a ? a 1 6a ? (22b )
>
>
>
>
1 b ? a 1 b ? (22b )
> >
> >
> >
5 6a ? a> 2> 12a ? b 1 a ? b
2 2b ? b
1
1
5 6(1) 2 12a b 1 a b
2
2
2 2(1)
3
52
2 > >
x?y
b. We have that cos (60°) 5 0 x> 0 0 y> 0 . Also since
>
>
0 x 0 5 3, 0 y 0 5 4, and cos (60°) 5 12,
> >
> >
>
x ? y 5 12 (4)(3) 5 6. Also x ? x 5 0 x 0 2 5 9
> >
>2
and y ? y 5 0 y 0 5 16.
The dot product is distributive, so
>
>
>
>
>
>
>
(4x 2 y ) ? (2x 1 3y ) 5 4x ? (2x 1 3y )
>
>
>
2 y ? (2x 1 3y )
> >
> >
> >
5 8x ? x 1 12x ? y 2 2y ? x
> >
2 3y ? y
5 8(9) 1 12(6) 2 2(6)
2 3(16)
5 84
29. The origin, (0, 0, 0), and (21, 3, 1) are two
points on this line. So (21, 3, 1) is a direction vector
for this line and since the origin is on the line, a
>
possible vector equation is r 5 t(21, 3, 1), tPR.
(21, 3, 1) is a normal vector for the plane. So the
equation of the plane is 2x 1 3y 1 z 1 D 5 0.
Calculus and Vectors Solutions Manual
(21, 3, 1) is a point on the plane. Substitute the
coordinates to determine the value of D.
119111D50
D 5 211
The equation of the plane is 2x 1 3y 1 z 2 11 5 0.
30. The plane is the right bisector joining
P(21, 0, 1) and its image. The line connecting the
two points has a direction vector equal to that of the
normal vector for the plane. The normal vector for
the plane is (0, 1, 21). So the line connecting the
two points is (21, 0, 1) 1 t(0, 1, 21), tPR, or in
corresponding
parametric form is x 5 21, y 5 t, z 5 1 2 t, tPR.
The intersection of this line and the plane is the
bisector between P and its image. To find this point
we plug the parametric equation into the plane
equation and solve for t.
0x 1 y 2 z 5 0(21) 1 (t) 2 (1 2 t)
5 21 1 2t
So if y 2 z 5 0, then 21 1 2t 5 0, or t 5 12.
So the point of intersection is occurs at t 5 12, since
the origin point is P and the intersection occurs at
the midpoint of the line connecting P and its image,
the image point occurs at t 5 2 3 12 5 1. So the
image point is at x 5 21, y 5 1, z 5 1 2 (1) 5 0.
So the image point is (21, 1, 0).
31. a. Thinking of the motorboat’s velocity vector
(without the influence of the current) as starting
at the origin and pointing northward toward the
opposite side of the river, the motorboat has velocity
vector (0, 10) and the river current has velocity
vector (4, 0). So the resultant velocity vector of the
motorboat is
(0, 10) 1 (4, 0) 5 (4, 10)
To reach the other side of the river, the motorboat
needs to cover a vertical distance of 2 km. So the
hypotenuse of the right triangle formed by the
marina, the motorboat’s initial position, and the
motorboat’s arrival point on the opposite side of
the river is represented by the vector
1
4
(4, 10) 5 a , 2b
5
5
(We multiplied by 15 to create a vertical component
of 2 in the motorboat’s resultant velocity vector,
the distance needed to cross the river.) Since this
new vector has horizontal component equal to 45,
this means that the motorboat arrives 45 5 0.8 km
downstream from the marina.
9-43
b. The motorboat is travelling at 10 km> h, and in
part a. we found that it will travel along the vector
( 45, 2) . The length of this vector is
cos u 5
4
4 2
` a , 2b ` 5 a b 1 22
5
Å 5
5 "4.64
So the motorboat travels a total of !4.64 km to
cross the river which, at 10 km> h, takes
"4.64 4 10 8 0.2 hours
5 12 minutes.
32. a. Answers may vary. For example:
A direction
vector for this line is
>
AB 5 (6, 3, 4) 2 (2, 21, 3)
5 (4, 4, 1)
So, since the point B(6, 3, 4) is on this line, the
vector equation of this line is
>
r 5 (6, 3, 4) 1 t(4, 4, 1), tPR.
The equivalent parametric form is
x 5 6 1 4t
y 5 3 1 4t
z 5 4 1 t, tPR.
b. The line found in part a. will lie in the plane
x 2 2y 1 4z 2 16 5 0 if and only if both points
A(2, 21, 3) and B(6, 3, 4) lie in this plane.
We verify this by substituting these points into the
equation of the plane, and checking for consistency.
For A:
2 2 2(21) 1 4(3) 2 16 5 0
For B:
6 2 2(3) 1 4(4) 2 16 5 0
Since both points lie on the plane, so does the line
found in part a.
33. The wind velocity vector is represented by (16, 0),
and the water current velocity vector is represented
by (0, 12). So the resultant of these two vectors is
(16, 0) 1 (0, 12) 5 (16, 12).
Thinking of this vector with tail at the origin and
head at point (16, 12), this vector forms a right
triangle with vertices at points (0, 0), (0, 12), and
(16, 12). Notice that
0 (16, 12) 0 5 "162 1 122
5 "400
5 20
This means that the sailboat is moving at a speed
of 20 km> h once we account for wind and water
velocities. Also the angle, u, this resultant vector
makes with the positive y-axis satisfies
9-44
12
20
u 5 cos 21 a
12
b
20
8 53.1°
So the sailboat is travelling in the direction
N 53.1° E, or equivalently E 36.9° N.
34. Think of the weight vector for the crane with tail
at the origin at head at (0, 2400) (we use one unit
for every kilogram of mass). We need to express this
weight vector as the sum of two vectors: one that is
parallel to the inclined plane and pointing down this
>
incline (call this vector x 5 (a, b)), and one that is
perpendicular to the inclined plane and pointing
>
toward the plane (call this vector y 5 (c, d)). The
>
angle between x and (0, 2400) is 60° and the angle
>
>
>
between y and (0, 2400) is 30°. Of course, x and y
are perpendicular. Using the formula for dot product,
we get
>
>
y ? (0, 2400) 5 0 y 0 0 (0, 2400) 0cos 30°
2400d 5 400a
"3
b"c 2 1 d 2
2
22d 5 "3 ? "c 2 1 d 2
4d 2 5 3(c 2 1 d 2 )
d 2 5 3c 2
So, since c is positive and d is negative (thinking of
the inclined plane as moving upward from left to
>
right as we look at it means that y points down and
d
to the right), this last equation means that c 5 2"3
>
So a vector in the same direction as y is (1, 2"3).
>
We can find the length of y by computing the scalar
projection of (0, 2400) on (1, 2 !3), which equals
(0, 2400) ? (1, 2"3)
400"3
5
2
0 (1, 2"3) 0
5 200"3
>
That is, 0 y 0 5 200"3. Now we can find the length
>
of x as well by using the fact that
>
>
0 x 0 2 1 0 y 0 2 5 0 (0, 2400) 0 2
>
0 x 0 2 1 (200"3)2 5 4002
>
0 x 0 5 "160 000 2 120 000
5 "40 000
5 200
Chapter 9: Relationships Between Points, Lines, and Planes
So we get that
>
>
0 x 0 5 200 and 0 y 0 5 200"3. This means that the
component of the weight of the mass parallel to the
inclined plane is
>
9.8 3 0 x 0 5 9.8 3 200
5 1960 N,
and the component of the weight of the mass
perpendicular to the inclined plane is
>
9.8 3 0 y 0 5 9.8 3 200"3
8 3394.82 N.
35. a. True; all non-parallel pairs of lines intersect
in exactly one point in R2. However, this is not
the case for lines in R3 (skew lines provide a
counterexample).
b. True; all non-parallel pairs of planes intersect in a
line in R3.
c. True; the line x 5 y 5 z has direction vector
(1, 1, 1), which is not perpendicular to the normal
vector (1, 22, 2) to the plane x 2 2y 1 2z 5 k,
k any constant. Since these vectors are not
perpendicular, the line is not parallel to the plane,
and so they will intersect in exactly one point.
d. False; a direction vector for the line
z11
x
5 y 2 1 5 2 is (2, 1, 2). A direction vector
2
x21
y21
z11
for the line 24 5 22 5 22 is (24, 22, 22),
or (2, 1, 1) (which is parallel to (24, 22, 22)).
Since (2, 1, 2) and (2, 1, 1) are obviously not
parallel, these two lines are not parallel.
36. a. A direction vector for
y22
L1: x 5 2,
5z
3
is (0, 3, 1), and a direction vector for
z 1 14
L2: x 5 y 1 k 5
k
is (1, 1, k). But (0, 3, 1) is not a nonzero scalar
multiple of (1, 1, k) for any k since the first
Calculus and Vectors Solutions Manual
component of (0, 3, 1) is 0. This means that the
direction vectors for L1 and L2 are never parallel,
which means that these lines are never parallel for
any k.
b. If L1 and L2 intersect, in particular their
x-coordinates will be equal at this intersection point.
But x 5 2 always in L1 so we get the equation
25y1k
y522k
y22
Also, from L1 we know that z 5 3 , so substituting
this in for z in L2 we get
2k 5 z 1 14
y22
2k 5
1 14
3
3(2k 2 14) 5 y 2 2
y 5 6k 2 40
So since we already know that y 5 2 2 k, we
now get
2 2 k 5 6k 2 40
7k 5 42
k56
So these two lines intersect when k 5 6. We have
already found that x 5 2 at this intersection point,
but now we know that
y 5 6k 2 40
5 6(6) 2 40
5 24
y22
z5
3
24 2 2
5
3
5 22
So the point of intersection of these two lines is
(2, 24, 22), and this occurs when k 5 6.
9-45