Math 3260
Solved problems in Chapter 3
Here are some hints and solved problems in Chapter 3 to help you in doing your homework.
First, some general hints in solving problems in Section 3.5.
A.
To determine whether two planes are parallel you have to check if their normal vectors are parallel (i.e. one is a constant multiple of the other).
B.
Two planes are perpendicular when their normal vectors are perpendicular. You check perpendicularity with the dot product.
C.
A line is perpendicular to a plane if it is parallel to the normal to the plane.
D.
A line is parallel to a plane if it is perpendicular to the normal vector to the plane.
E.
A line given in parametric equation is contained in a plane if when substituting x , y and z in the equation of the plane you get an identity involving the parameter t .
F.
To find the intersection of a line given in parametric equation with a plane, you have to solve for the parameter t the equation obtained when substituting x , y and z in the equation of the plane. If no such t exists, the line does not intersect the plane, therefore it is parallel to it. After you find the value of t , find the coordinates of the intersection point from the parametric equations.
Section 3.3. Ex.11. Denote A(0, -1), B(1, -2), C(4, 1). Write the sides of the triangle as vectors. Let’s define u

A B

( 1 ,

1 ), v

A C

( 4 , 2 ), w

B C

( 3 , 3 ) . Use the formula for the dot product to find the cosine of the interior angles. We get cos( A )
 u u

 v v

2

2
20

1
10
, and similarly for the other angles.
Ex. 12 . Proceed as in Ex.11 and look for the two vectors that are orthogonal, i.e. their dot product is zero.
Ex . 13 . We try to find w

( x , y , z ), such that w
 u

0 , w
 v

0 , w

1 . We get x y

 z z


0
0
so w

(
 t ,
 t , t ) and since w

1 , 3 t
2 
1 , t
  1
3
. So we get two vectors: w
1

1
3
( 1 , 1 ,

1 ), w
2

1
3
(

1 ,

1 , 1 ) .
Ex. 16 . a) The vectors are parallel if q
 t
 p for some value of t , from which we get t b)

3
2
, and therefore k p
 q

0 so k
 
6
5
.

10
3
.

d) We know that the cosine of the angle between the vectors is
1
2

6
34


5 k
4
 k
2
.
Square both sides and multiply by the denominator to obtain:
136

34 k
2 
2 ( 36

60 k

25 k
2
) . Solve the quadratic equation to get k

1
2
and k
Since we squared the equation we will check if both are solutions to our problem:
 
8 .
For k

1
2
we get the cosine of the angle solution is k

1
2
.
Ex. 19.
2
2
, and for k
 
8 we get

2
2
, so the only
1
4
( u+v
2   2 v )

1
4
     v)]
.
1
2
= [ u 2u v+ v
4
  2  u
2    2 v ]
 
Ex. 24 . (a) We will find the area of the triangle using the well-known formula base
 height
Area

. We will take CD as base and find the length of the perpendicular
2 from A onto CD. But that perpendicular is the vector component of CA orthogonal to CD.
We have
CA

(

2 ,

4 ), CD

(

9 , 1 ), proj
CD
CA

14
82
(

9 , 1 ), CA
 proj
CD
CA

1
41
(

19 ,

157 )
So the area is
CD

CA
 proj
CD
CA
2

82
25010
41
2
2

17 .
46 .

Now, to complete the triangle (see above) to a parallelogram, we have several options for the fourth vertex B. We can find B such that BD//AC, BD=AC (two positions, one below the point D and one above), or such that AB//CD, AB=CD) (this will give another two positions, one to the left of A which coincides with a solution that we find above, and another one to the right of A. Therefore we have:
BD=AC (as vectors) will give B(-7,4)
DB=AC (as vectors) will give B(-3,12)
AB=DC (as vectors) will give B(11, 2).
As for the area of the parallelogram, it is twice the area of the triangle.
Ex. 25 . For any scalars k
1
, k
2
, v

( k
1 w
1
 k
2 w
2
)
 k
1 v
 w
1
 k
2 v
 w
2

0 which proves the statement.
Section 3.5 Problem 23 .
The vector normal to the plane 2 x

3 y

4 z 7 0 is n

(2, 3, 4).
We need a plane perpendicular onto this plane so its normal vector,
We have 2 a

4 b

2 c

0, therefore n
1

(2 b c b c n
1
satisfies n n
1
0 . Let n
1
 a b c .
, with b and c to be determined from the condition that the plane contains the given line.
Notice that from the parametric equations of the line, we see that one point on the line is
P

0
( 1,5, 2) . Let us write the equation of the plane as n P P
1 0

0, which is
(2

)(
  b y
  c z

2)

0 . Now since the given line is contained in the plane, when we substitute the expressions for x , y and z from the parametric equations we must obtain an identity in t . We get: 3(2 )

2 bt ct 0 . Since this must be zero for

all values of t , it follows that 3(2 b c ) 2 b c 0, or 2
 
0, therefore c

2 b . So we obtain the normal to the plane of the form value for b , so let b

1, n
1
 n
1

. We can choose any particular
(0,1, 2), and the plane is y

2 z
 
0 .
Problem 26 . Since the line is parallel to both planes (being parallel to their line of intersection), the direction vector of the line we have to find is perpendicular to both normal vectors to the given planes. Let v

( , , ) be the direction vector of the line.
Then v (2,1, 4) 0, and v ( 1, 2, 3)

0.
Therefore we have to solve the system:


2 a a

2 b b

4
3 c c

0
0
. Solve this system (which has infinitely many solutions), and set c t 5 , to get a

11, b
 
2 . Therefore the line has the parametric equations x
     t z
 t
Problem 31 . Let the equation of the plane be of the form ax by cz d 0 . Since the point P (1, 1, 2) and the line x

,
 
1, z 3 2 t
 
2 c
 
0, (
 
2 )
 
3 c
 
0
are in the plane, we get:
. Now, since the second equation is satisfied for all values of t , it must be that
 
2 c

0, and b

3 c d 0 . Therefore we have the system:


 a b b
2 c d a b 2 c

  
0
0
0
. One solution to this system is a
 
3, b c 1, d

2 , and therefore the plane has equation 3 x y z 2 0 .
Problem 34 . In order to check that the line is parallel to the plane, it will be sufficient to check that the direction vector of the line is perpendicular to the normal to the plane. We have v ( 1, 2, 5), n ( 3,1,1), and v n=0 , therefore the line is parallel to the plane.
Problem 40 a) Since the planes are parallel, the distance between the two planes is the same as the distance from any arbitrary point in one plane to the other plane. Therefore, let us choose a point in 3 x

4 y z 1 . Take for example P
0
(0, 0,1) . We now use the distance formula in Theorem 3.5.2: D

 

1
36 64 4 104
.