United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 1 – SOLUTION
Section 10.1 Vectors in the Plane
Calculus II for Engineering
MATH 1120 SECTION 04 CRN 23510
2:00 – 4:00 on Monday & Wednesday
Due Date: Wednesday, October 6, 2010
Calculus II for Engineering
1. Compute
HOMEWORK 1 – SOLUTION
Fall, 2010
3a + 2b for a = h 3; 2 i and b = h 3; 13 i.
Answer.
3a + 2b = 3 h 3; 2 i + 2 h 3; 13 i = h 9; 6 i + h 6; 26 i
= h 9 + 6; 6 26 i = h 3; 32 i :
2. Determine whether the vectors a = h 1;
2 i and b = h 2; 1 i are parallel.
Answer. We recall that a and b are parallel if and only if there is a scalar s such that a = sb. If a = sb,
then
h 1; 2 i = s h 2; 1 i = h 2s; s i ; 1 = 2s and 2 = s:
There is no such a scalar s satisfying both 2s
parallel.
= 1 and s = 2. Therefore, given two vectors cannot be
3. Find the vector with initial point A(2; 3) and terminal point B (5; 4).
! = h 5 2; 4 3 i = h 3; 1 i :
Answer. AB
4. Find a vector with the magnitude 4 in the same direction as the vector v = 2i
j.
Answer. Let u be such a vector, i.e., the one with the magnitude 4 in the same direction as the vector
v = 2i j. Since u is parallel to v, there should be a positive scalar s (negative scalar gives the opposite
direction) such that u = sv. Since u has the magnitude 4, we have
p
4 = kuk = ksvk = jsjkvk = jsjk2i j k = jsj 5;
p
p
4
4
4
5
5:
;
i:e:; s =
i:e:; jsj = p =
5
5 5
p
4
5 (2i j ) :
Therefore, we deduce such a vector u =
5
Page 1 of 4
Calculus II for Engineering
HOMEWORK 1 – SOLUTION
Fall, 2010
5. The thrust of an airplane’s engines produces a speed of 600 mph in still air. The wind velocity is given
by h
30; 60 i. In what direction should the airplane head to fly due east?
Answer. Let v = h x; y i and w = h 30; 60 i be the velocities of the airplane and the wind, respectively. Since we want the airplane to move due east, the sum of two vectors v and w should satisfy
v + w = h c; 0 i, where the vector h c; 0 i points due east with the positive constant c. The equation implies
h c; 0 i = v + w = h x; y i + h 30; 60 i = h x 30; y + 60 i ;
i:e:; c = x 30; and y = 60:
The airplane produces a speed of 600 mph and so we get kv k = 600, which implies
q
600 = kvk = h x; y i = x2 + y2;
p
p
x = 6002 602 = 180 11:
6002 = x2 + y2 = x2 + ( 60)2;
p
30 = c > 0, we choose x = +180 11, hence,
D p
E
D p
E
v = h x; y i = 180 11; 60 = 60 3 11; 1 :
Because of the condition that x
This points right and down at an angle of tan
east.
1
!
p1 5:73917 0:100167 (radian) south of
3 11
6. Suppose a vector a has magnitude kak = 3 and vector b has magnitude kbk = 4.
(6.1) What is the largest possible magnitude for the vector a + b?
Answer. The largest magnitude of a + b is 7 (if the vectors point in the same direction).
(6.2) What is the smallest possible magnitude for the vector a + b?
Answer. The smallest magnitude is 1 (if the vectors point in the opposite directions).
(6.3) What will be the magnitude of a + b if a and b are perpendicular?
Answer. If the vectors are perpendicular, then a + b can be viewed as the hypotenuse of a right
p
triangle with sides a and b, so it has length 32 + 42 = 5.
Page 2 of 4
Calculus II for Engineering
7. A vector 4i
HOMEWORK 1 – SOLUTION
Fall, 2010
2j is given. Here i = h 1; 0 i and j = h 0; 1 i.
(7.1) Find two unit vectors parallel to the given vector.
Answer. Let v
= 4i 2j = 2 (2i
kvk = k2 (2i
j ). Then it has the magnitude
j ) k = 2k (2i
j) k = 2
p
22 + ( 1)2 = 2 5:
q
The unit vector in the same direction as v can be found by
v
kv k
= 2 (22ip5 j ) = p15 (2i
j) :
The unit vector in the opposite direction as v is
= p15 (2i
1
Thus, two unit vectors parallel to v are p (2i j ).
5
v
kv k
j) :
(7.2) Write the given vector as the product of its magnitude and a unit vector.
Answer. Using the results above, we have
v
v = kv k
kvk
p
= 2 5 p25 i p15 j
!
:
8. If v 2 V2 lies in the first quadrant of the xy –plane and makes the angle = =3 with the positive x–axis
and the magnitude 4, then find v in the component form.
Answer. The unit vector making the angle = =3 with the positive x–axis is given by h cos(=3); sin(=3) i.
(Why?) Hence, the desired vector v 2 V2 is obtained by
p + D p E
1
3 = 2; 2 3 :
v = 4 h cos(=3); sin(=3) i = 4
;
2 2
*
Page 3 of 4
Calculus II for Engineering
HOMEWORK 1 – SOLUTION
Fall, 2010
= h 3; 2 i, b = h 2; 1 i and c = h 7; 1 i. By using the sketch involved
with those vectors, show that there exist scalars s and t such that c = sa + tb. Can you prove the
9. (Think!) Draw the vectors a
existence of such s and t algebraically? Justify your answer.
Answer. The equation c = sa+tb implies h 7; 1 i = s h 3; 2 i + t h 2;
9
11
Solving the equations for s and t, we get s = and t = .
7
7
10. Find the correct figure of the sum a + b with a =
2i
j and b =
1 i, i.e., 7 = 3s+2t and 1 = 2s
t.
3i +2j, where i and j are standard
basis vectors of V2 .
Answer. A simple computation shows
a+b=
2i
j
3i + 2j = 5i + j = h 5; 1 i :
That is, a + b should point in the direction h 5; 1 i. It is easy to see that the red–colored vector in (2)
is the vector h 5; 1 i. Hence, the answer is (2).
H1L
5
-5
-5
-4
-4
-3
-3
-2
-2
H2 L
5
4
4
3
3
2
2
1
1
1
-1
2
3
4
5
-5
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
H3 L
5
H4L
5
4
4
3
3
2
2
1
1
1
-1
2
3
4
5
-5
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
Page 4 of 4
2
2
3
3
4
4
5
5
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 2 – SOLUTION
Section 10.2 Vectors in the Space
Calculus II for Engineering
MATH 1120 SECTION 04 CRN 23510
2:00 – 4:00 on Monday & Wednesday
Due Date: Wednesday, October 13, 2010
Calculus II for Engineering
HOMEWORK 2 – SOLUTION
1. Find two unit vectors parallel to the vector 4i
Answer. Let v
= 4i 2j + 4k = 2 (2i
kvk = k2 (2i
Fall, 2010
2j + 4k.
j + 2k). Then it has the magnitude
j + 2k) k = 2k2i
j + 2k k = 2
q
p
22 + ( 1)2 + 22 = 2 9 = 6
:
The unit vector in the same direction as v can be found by
v
kv k
= 2 (2i 6j + 2k) = 13 (2i
j + 2k ) :
The unit vector in the opposite direction as v is
v
kvk
Thus, two unit vectors parallel to v are = 13 (2i
1 (2i
3
j + 2k ) :
j + 2 k ).
2. Identify the plane y = 4 as parallel to the xy –plane, xz –plane or yz –plane and sketch a graph.
Answer. The plane y
= 4 is parallel to the
!
xz
–plane and passes through (0; 4; 0).
!
3. Find the displacement vectors P Q and QR and determine whether the points P (2; 3; 1), Q(0; 4; 2) and
R
(4 1 4) are colinear (on the same line).
;
;
Answer. The displacement vectors are
!=h0 2 4 3 2 1i=h 2 1 1i
!=h4 0 1 4 4 2i=h4 3 2i
! = !, because no scalar satisfies simultaneously
There does not exist any scalar such that
PQ
;
;
;
s
;
;
PQ
QR
;
sQR
It implies that two vectors P Q
s;
QR
Page 1 of 3
;
;
:
s
2=4
1= 3
1=2
! and ! are not parallel. That is, the points
s;
;
s:
P
, Q and R are not colinear.
Calculus II for Engineering
HOMEWORK 2 – SOLUTION
4. Use vectors to determine whether the points (2; 1; 0), (5;
Answer. We recall that if a square has the side length s
Pythagorean Theorem.
Fall, 2010
1 2), (0 3 3) and (3 1 5) form a square.
p
, then it has the diagonal of length
2 by the
;
;
;
;
;
s
!
Let P = (2; 1; 0), Q = (5; 1; 2), R = (0; 3; 3) and S = (3; 1; 5). There are six pairs of vectors (P Q,
! ! ! ! !
! !
!
P R, P S , QR, QS , RS ) and four of them will correspond to sides. Further P Q, P R and P S should
form two sides and one diagonal of the square, because they have the same initial point. When we
compute the lengths, we get
p
k !k = k h 3 2 2 i k = 17
p
k !k = k h 1 0 5 i k = 26
PQ
;
PS
;
;
p
k !k = k h 2 2 3 i k = 17
;
;
PR
;
;
;
:
p
!
p p
! p2, so !
!
It implies P S should be the diagonal. However, since 26 6= 17 2, i.e., kP S k 6= kP Qk
cannot be the diagonal of a square. That is, those given points cannot form a square.
PS
5. In the accompanying figure, two ropes are attached to a 300–pound crate. Rope A exerts a force of
h 10 130 200 i pounds on the crate, and rope
;
;
B
exerts a force of
h 20 180 160 i pounds on the
;
;
crate.
(5.1) If no further ropes are added, find the net force on the crate and the direction it will move.
Answer. Let the force due to rope A be a = h 10; 130; 200 i, the force due to rope B be b =
h 20; 180; 160 i, and write the force due to gravity as w = h 0; 0; 300 i. Then the net force is
a+b+w
= h 10 130 200 i + h 20 180 160 i + h 0 0 300 i = h 10 50 60 i
;
;
;
;
;
;
;
;
:
(5.2) If a third rope C is added to balance the crate, what force must this rope exert on the crate?
Answer. In order to compensate, rope C must exert a force of h 10;
78.74 (= k h 10;
50 60 i k) pounds in direction h 1 5 6 i.
;
;
Page 2 of 3
;
50 60 i or
;
Calculus II for Engineering
HOMEWORK 2 – SOLUTION
Fall, 2010
(5.3) We want to move the crate up and to the right with a constant force of h 0; 30; 20 i pounds. If a
third rope C is added to accomplish this, what force must the rope exert on the crate?
Answer. Let the force due to rope C be c. We want the net force to be
a+b+c+w
= h 0 30 20 i
;
;
;
c+h
i:e:;
10 50 60 i = h 0 30 20 i
;
;
;
;
;
where a, b and w defined in (1) above are used. That is,
c = h 0; 30; 20 i
So rope C must exert a force of h 10;
tion h 1; 2; 4 i.
h 10 50 60 i = h 10 20 40 i
;
;
;
;
:
20 40 i or 45.8 (= k h 10 20 40 i k) pounds in direc;
Page 3 of 3
;
;
United Arab Emirates University
Faculty of Science
Department of Mathematical Sciences
HOMEWORK 3 – SOLUTION
Section 10.3 Dot Product and Section 10.4 Cross Product
Calculus II for Engineering
MATH 1120 SECTION 04 CRN 23510
2:00 – 4:00 on Monday & Wednesday
Due Date: Wednesday, October 20, 2010
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Section 10.3 Dot Product
1. Compute a b.
(1.1) a = h 3; 2; 0 i and b = h
Answer.
(1.2) a = 2i
2; 4; 3 i .
a b = h 3; 2; 0 i h
k and b = 4j
Answer.
2; 4; 3 i = 3( 2) + 2(4) + 0(3) = 2:
k.
a b = (2i
k) (4j
k) = 2(0) + 0(4) + (
1)( 1) = 1:
2. Compute the angle between the vectors.
(2.1) a = h 2; 0;
2 i and b = h 0; 2; 4 i.
Answer.
cos = kaakk bbk = k hh22;;00;; 22iikk hh00;; 22;;44ii k = p210 ;
p2 2:25552
= cos 1
10
!
(2.2) a = 3i + j
4k and b = 2i + 2j + k.
Answer.
cos = kaakk bbk = (3k3ii++jj 44kkk) ( 22ii++22jj++kkk) = 3p826 ;
p8 2:12114
= cos 1
3 26
!
3. Determine whether the vectors are orthogonal.
(3.1) a = h 4;
1; 1 i and b = h 2; 4; 4 i.
Answer.
a b = h 4;
1; 1 i h 2; 4; 4 i = 8 6= 0:
So a and b are not orthogonal.
(3.2) a = 6i + 2j and b =
i + 3j.
Answer.
a b = (6i + 2j ) ( i + 3j ) = 0:
So a and b are orthogonal.
4. Find a vector perpendicular to the given vector.
Page 1 of 9
Calculus II for Engineering
(4.1)
HOMEWORK 3 – SOLUTION
Fall, 2010
h 4; 1; 1 i .
Answer. Let v
v u = 0, i.e.,
= h a; b; c i be a vector perpendicular to u = h 4; 1; 1 i.
0 = v u = h a; b; c i h 4; 1; 1 i = 4a
b
+ c;
4a
i:e:;
Then we should have
b
+ c = 0:
There are so many numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = 4
and c = 0. That is, v = h 1; 4; 0 i is one vector perpendicular to u = h 4; 1; 1 i.
(4.2) 6i + 2j
k.
Answer. Let v = h a; b; c i = ai + bj + ck be a vector perpendicular to w
by the same argument as above, we deduce
0 = v w = (ai + bj + ck) (6i + 2j
k ) = 6 a + 2b
c;
i:e:;
= 6i + 2 j
6a + 2 b
k. Then
c
= 0:
There are so many numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = 1
and c = 4. That is, v = h 1; 1; 4 i = i j + 4k is one vector perpendicular to w = 6i + 2j
k.
5. Find Compb a and Projb a.
(5.1) a = 3i + j and b = 4i
3j.
Answer.
= (3i +k4ji) (43ij k 3j ) = 95
b
9 4i 3j = 9 (4i 3j )
Projb a = (Compb a)
=
kbk 5 k4i 3j k 25
Compb a =
(5.2) a = h 3; 2; 0 i and b = h
ab
kbk
2; 2; 1 i .
Answer.
= h 3; k2;h0 i2;h2; 12i; 2k; 1 i = 23
b
2 h 2; 2; 1 i = 2 h 2; 2; 1 i
Projb a = (Compb a)
=
kbk 3 k h 2; 2; 1 i k 9
Compb a =
6. A constant force of h 60;
point (10;
ab
kbk
30 i pounds moves an object in a straight line from the point (0; 0) to the
10). Compute the work done.
Answer. The displacement vector is d = h 10 0;
By the formula, the work W done is obtained by
W
10 0 i = h 10; 10 i. The force is F = h 60; 30 i.
= F d = h 10; 10 i h 60; 30 i = 10(60) + ( 10)( 30) = 900:
Page 2 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
7. Label each statement as true or false. If it is true, briefly explain why; if it is false, give a counterexample.
(7.1) If a b = a c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. a = h 1; 0; 0 i, b = h 0; 1; 0 i and c = h 0; 0; 1 i satisfies a b = 0 = a c. But, obviously,
b 6= c.
(7.2) If b = c, then a b = a c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. If b = c, then b
a (b
T
c = 0 and so
c) = a 0 = 0 ;
i:e:;
ab
a c = 0;
i:e:;
a b = a c:
(7.3) a a = kak2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. The formula a b = kakkbk cos , where is the angle between a and b, implies
a a = kakkak cos 0 = kak2 ;
i:e:;
a a = kak2 :
One may compute the dot product with a = h a; b; c i and prove the equality.
(7.4) If kak > kbk, then a c > b c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. We observe a = h 2; 0 i and b = h 0; 1 i satisfy the inequality kak
However, with c = h 0; 3 i, we get a c = 0 < 3 = b c.
= 2 > 1 = kbk.
(7.5) If kak = kbk, then a = b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. We observe i = h 1; 0; 0 i, j
However, obviously, i 6= j 6= k.
F
F
= h 0; 1; 0 i and k = h 0; 0; 1 i have kik = 1 = kj k = kkk.
8. By the Cauchy–Schwartz Inequality, ja bj kakkbk. What relationship must exist between a and b
to have the equality ja bj = kakkbk?
Answer. (1) a = 0 or b = 0. (2) The cosine of the angle between the vectors is 1. This happens
exactly when the vectors point in the same or opposite directions. In other words, when a = sb for
some scalar s.
9. By the Triangle Inequality, ka + bk kak + kbk. What relationship must exist between a and b to
have the equality ka + bk = kak + kbk?
Answer. (1) a = 0 or b = 0. (2) Vectors a and b must be parallel with the same direction so that
a = sb for some positive scalar s.
10. The orthogonal projection of vector a along vector b is defined as Orthb a = a
Projb a. Sketch a
picture showing vectors a, b, Projb a and Orthb a, and explain what is orthogonal about Orthb a.
Page 3 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Answer. Orthb a is the component of a that is orthogonal to b:
b Orthb a = b (a
=ba
Projb a) = b a
b
b Projb a
(a b)b = b a (a b)(b b) = b a
kbk2
kbk2
where b b = kbk2 is used above.
a b = 0;
11. A car makes a turn on a banked road. If the road is banked at 15 , show that a vector parallel to the
road is h cos 15 ; sin 15 i. If the car has weight 2500 pounds, find the component of the weight vector
along the road vector. This component of weight provides a force that helps the car turn.
Answer. The vector b = h cos 15 ; sin 15 i represents the direction of the banked road. The weight of
the car is w = h 0; 2500 i. The component of the weight in the direction of the bank is
Compb w
= wkbkb = 2500 sin 15 647:0 lbs
toward the inside of the curve.
Page 4 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Section 10.4 Cross Product
12. Compute the determinant:
Answer.
0 2 1
1 1 2
1 1 2
0 2 1
1 1 2 = 0( 2 2) 2(2 2) (1 + 1) = 2:
1 1 2
13. Compute the cross product a b.
(13.1) a = h 2;
2; 0 i and b = h 3; 0; 1 i.
Answer.
a b = h 2;
(13.2) a =
i
j
k
2; 0 i h 3; 0; 1 i = 2 2 0 = 2i 2j + 6k = 2 h 1; 1; 3 i 3 0 1
2i + j 3k and b = 2j
k.
Answer.
a b = h 2; 1;
3 i h 0; 2; 1 i =
i
j
2 1
0 2
k
3 = 5i 2 j 4 k = h 5 ; 2 ; 4 i 1
14. Find two unit vectors orthogonal to the two given vectors.
(14.1) a = h 0; 2; 1 i and b = h 1; 0;
1 i.
Answer. We recall
Orthogonal Vector
a b is orthogonal to both a and b. Thus, a and b.
a b = h 0; 2; 1 i h 1; 0;
(14.2) a =
2i + 3j 3k and b = 2i
ab
k a b k are two unit vectors orthogonal to both
1 i = h 2; 1; 2 i ; k aa bb k = 13 h 2; 1; 2 i
k.
Answer.
ab=h
1 h 3; 8; 6 i
2; 3; 3 i h 2; 0; 1 i = h 3; 8; 6 i ; k aa bb k = p109
Page 5 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
15. Use the cross product to determine the angle between the vectors, assuming that 0 =2.
(15.1) a = h 2; 2; 1 i and b = h 0; 0; 2 i.
Answer. We recall
Angle Between Two Vectors
k a b k = k a k k b k sin ;
where 0 i:e:;
sin = kkaakkbbkk ;
i:e:;
is the angle between a and b.
a b = 4 h 1;
= sin
= sin
1
kabk
k a kk b k
p
1; 0 i ; k a k = 3; k b k = 2; k a b k = 4 2;
p
4 2 1:23096 rad 70:5288
6
!
1
!
(15.2) a = i + 3j + 3k and b = 2i + j.
Answer.
p
p
p
3; 6; 5 i ; k a k = 19; k b k = 5; k a b k = 70;
p
70 1:03213 rad 59:1369
= sin 1 p
95
ab=h
!
16. Find the distance from the point Q(1; 3; 1) to the line through (1; 3;
2) and (1; 0; 2).
Answer. The distance d from the point Q to the line through the points P and R is obtained by the
following formula.
Distance from a Point to a Line
d
=
! P!R
!
PR
PQ
= (1; 3; 2) and R = (1; 0; 2), we have
!
!
! !
P Q = 3 h 0; 0; 1 i ; P R = 3 h 0; 1; 0 i ; P Q P R = 9 h 1; 0; 0 i ;
Letting P
d
= 39 kk hh 01;; 10;; 00 ii kk = 3: 17. If you apply a force of magnitude 30 pounds at the end of an 8–inch–long wrench at an angle of =3 to
the wrench, find the magnitude of the torque applied to the bolt.
Answer. We recall
Page 6 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Torque
Torque τ is defined to be the cross product of the position vector r and force vector F , i.e.,
τ
= r F;
k τ k = k r k k F k sin where is the angle between r and F .
Given information: k F
k = 30 and = =3 and k r k = 8 inch = 8=12 = 2=3 feet. Thus we have
p
k τ k = 32 (30) sin 3 = 10 3 ft–lbs
18. Label each statement as TRUE or FALSE. If it is true, briefly explain why. If it is false, give a counterexample.
(18.1) If a b = a c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. We observe that for a = h 1; 0; 0 i, b = h 1; 1; 1 i and c = h 2; 1; 1 i,
a b = h 0;
(18.2) a b =
1 ; 1 i = a c;
but
b 6= c:
b a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. We observe that for a = h a1 ; a2 ; a3 i and b = h b1 ; b2 ; b3 i,
a b = h a2 b3
a3 b2 ; a1 b3
a3 b1 ; a1 b2
a2 b1
i=
b a:
(18.3) a a = kak2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. a a = h 0; 0; 0 i is a vector, while k a k is a scalar. As a counterexample, for a = i, we
get i i = 0 = h 0; 0; 0 i 6= 1 = k i k2 .
(18.4) a (b c) = (a b) c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. (a b) c is not possible because a b is a scalar. A cross product must involve two
vectors.
(18.5) If the force is doubled, the torque doubles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. Torque is the cross product of direction and force.
r (2F ) = 2(r F ) = 2τ :
19. Find the area of the parallelogram with two adjacent sides formed by h
Answer. We recall
Page 7 of 9
2; 1 i and h 1; 3 i.
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Area
The area of the parallelogram with two adjacent sides formed by a and b is the magnitude of their
cross product:
A = k a b k = k a k k b k sin where 0 is the angle between a and b.
The vectors h 2; 1 i and h 1; 3 i are in the plane (R2 ). The cross product is defined only for the
vectors in the space (R3 ). But we observe h 2; 1; 0 i and h 1; 3; 0 i in the space can correspond to
those vectors in the plane, respectively. So we compute the area with these vectors:
A
= k h 2; 1; 0 i h 1; 3; 0 i k = k 5 h 0; 0; 1 i k = 5:
20. Find the area of the triangle with vertices (0; 0; 0), (0;
2; 1) and (1; 3; 0).
Answer. Letting P (0; 0; 0), Q(0; 2; 1) and R(1; 3; 0), by the formula, the area of the parallelogram
!
!
with two adjacent sides formed by P Q and P R is the magnitude of their cross product:
Aparallelogram
p
!
= P!
Q P R = k h 0; 2; 1 i h 1; 3; 0 i k = k h 3; 2; 1 i k = 14:
Since the area of the triangle is the half of the area of the parallelogram, hence the desired area is
Atriangle
=
! P!R p
14 :
=
2
2
PQ
21. Find the volume of the parallelepiped with three adjacent edges formed by h 0;
h 1; 0; 2 i.
1; 0 i, h 0; 2; 1 i, and
Answer. We recall
Volume
The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their
scalar triple product:
V = jc (a b)j
Letting a = h 0;
1; 0 i, b = h 0; 2; 1 i and c = h 1; 0; 2 i, the formula implies the volume V = 1. 22. Use geometry to identify the cross product. (Do not compute!)
(22.1) j (j k).
Answer. Geometry implies
Page 8 of 9
Calculus II for Engineering
HOMEWORK 3 – SOLUTION
Fall, 2010
Cross Product on Standard Basis Vectors
=k
ji= k
ij
Hence, we deduce
jk =i
kj
=
ki=j
ik =
i
j (j k ) = j i =
j
k
kk =0
(22.2) (j i) k.
Answer. By the same argument as above, we get
( j i) k =
23. Use the parallelepiped volume formula to determine whether the vectors h 1; 1; 2 i and h 0;
h 3; 2; 4 i are coplanar.
1; 0 i and
Answer. Letting a = h 1; 1; 2 i, b = h 0; 1; 0 i and c = h 3; 2; 4 i, the formula above implies the
volume V = 2. Since the volume of the parallelepiped is not zero, it means those three vectors do not
lie on the same plane. That is, they are not coplanar.
Page 9 of 9
United Arab Emirates University
Faculty of Science
Department of Mathematical Sciences
HOMEWORK 4 – SOLUTION
Section 10.5 Lines and Planes in Space
Calculus II for Engineering
MATH 1120 SECTION 04 CRN 23510
2:00 – 4:00 on Monday & Wednesday
Due Date: Wednesday, October 27, 2010
Calculus II for Engineering
HOMEWORK 4 – SOLUTION
Fall, 2010
1. Find (a) parametric equations and (b) symmetric equations of the line.
(1.1) The line through (3;
2; 4) and parallel to h 3; 2; 1 i.
Answer.
(a)
x
= 3 + 3t;
y
= 2 + 2t;
=4
z
x
( b)
t
3 = y+2 = z 4 3
2
1
1; 0; 0) and parallel to the line x +21 = y3 = z 2.
The direction comes from the given line, i.e., h 2; 3; 1 i.
(1.2) The line through (
Answer.
(a)
x
= 1 2t;
y
= 0 + 3t;
z
=0+t
( b)
x
+1 = y = z
2 3 1
(1.3) The line through (
(1.4)
3; 1; 0) and perpendicular to both h 0; 3; 1 i and h 4; 2; 1 i.
Answer. h 0; 3; 1 i h 4; 2; 1 i = h 1; 4; 12 i is in the direction perpendicular to both vectors.
(a) x = 3 + t; y = 1 + 4t; z = 0 + 12t
(b) x +1 3 = y 4 1 = 12z The line through (0; 2; 1) and normal to the plane y + 3z = 4.
Answer. h 0; 1; 3 i is normal to the plane.
(a) x = 0 + 0t = 0; y = 2 + t; z = 1 + 3t
(b) x = 0; y +1 2 = z 3 1 2. State whether the lines are parallel or perpendicular and find the angle between the lines:
: x = 4 2t;
M : x = 4 + s;
L
= 3t; z = 1 + 2t
y = 2s; z = 1 + 3s:
y
Answer. The vectors parallel to lines L and M are respectively vL = h 2; 3; 2 i and vM = h 1; 2; 3 i.
Since there is no constant c satisfying vL = h 2; 3; 2 i = c h 1; 2; 3 i = cvM , so the lines L and M are
not parallel.
By the formula on the dot product, we have
cos =
vL vM
k vL k k vM
2 p6 + 6 = p 2 ;
p
=
k
17 14
238
= cos
1
!
p 2 1:7 rad;
238
which is not =2. Thus, the lines L and M are not perpendicular and the angle between the lines L and
M is about 1.7 rad.
3. Determine whether the lines are parallel, skew or intersect:
L
: x = 3 + t;
y
= 3 + 3t;
Page 1 of 5
z
=4
t
Calculus II for Engineering
HOMEWORK 4 – SOLUTION
M
: x=2
s;
y
= 1 2s;
Fall, 2010
z
= 6 + 2s:
Answer. The vectors parallel to lines L and M are respectively vL = h 1; 3; 1 i and vM = h 1; 2; 2 i.
Since there is no constant c satisfying vL = h 1; 3; 1 i = c h 1; 2; 2 i = cvM , so the lines L and M
are not parallel. To determine whether or not the lines intersect, we set the x–, y – and z –values equal:
simply;
3 + t = 2 s;
( i) t + s = 1 ;
3 + 3t = 1 2s;
(ii) 3t + 2s = 2;
4 t = 6 + 2s;
(iii) t + 2s = 2:
From (i) and (ii), we get t = 0 and s = 1. Putting t = 0 and s = 1 into (iii), the equation (iii)
holds. This implies that the lines L and M intersect when t = 0 and s = 1:
x
= 3;
y
= 3;
z
= 4:
Hence, the lines intersect at the point (3; 3; 4).
4. Find an equation of the given plane.
(4.1) The plane containing the point (
2; 1; 0) with normal vector h 3; 0; 2 i.
Answer.
3(x + 2) + 0(y 1) + 2(z 0) = 0;
3x 2z + 6 = 0 :
simply;
(4.2) The plane containing the points (
2; 2; 0), ( 2; 3; 2) and (1; 2; 2).
!
!
Answer. Letting P ( 2; 2; 0), Q( 2; 3; 2) and R(1; 2; 2), we deduce P Q = h 0; 1; 2 i and P R =
!
h 3; 0; 2 i and P!
Q P R = h 2; 6; 3 i is normal to the plane. Hence, using the point P ( 2; 2; 0)
(one can use Q or R), the equation of the plane is obtained as
2(x + 2) + 6(y 2) 3(z 0) = 0;
simply;
2x + 6 y 3z 8 = 0 :
(4.3) The plane containing the point (3;
Answer.
2; 1) and parallel to the plane x + 3y 4z = 2.
The vector normal to the plane is h 1; 3; 4 i. Hence, the equation of the plane is
(x 3) + 3(y + 2) 4(z 1) = 0;
(4.4) The plane containing the point (3; 0;
2 x z = 1.
simply;
x
+ 3 y 4z + 7 = 0 :
1) and perpendicular to the planes x + 2y
z
= 2 and
Answer. Let v1 and v2 be the vectors normal to the planes: v1 = h 1; 2; 1 i and v2 = h 2; 0; 1 i.
Normal vector must be perpendicular to the normal vectors of both planes. So, the plane that we
are looking for has the normal vector v1 v2 = h 2; 1; 4 i. Hence, the equation of the plane is
2(x 3) (y 0) 4(z + 1) = 0;
Page 2 of 5
simply;
2 x + y + 4 z = 2:
Calculus II for Engineering
HOMEWORK 4 – SOLUTION
Fall, 2010
5. Sketch the given plane.
(5.1) 2x
y
+ 4z = 4 .
Answer. The equation represents a plane which has the normal vector h 2; 1; 4 i and passes through
the point P (2; 0; 0), Q(0; 4; 0) and R(0; 0; 1). We connect those three points and sketch the plane
roughly.
Y-axis
4
2
0
-2
Normal Vector
-4
4
2
R
P
Z-axis 0
Q
-2
-4
-4
-2
0
2
X-axis
(5.2) x + y
4
= 1.
Answer. The equation represents a plane such that
(i) it is parallel to the z –axis and perpendicular to the xy–plane,
(ii) it has the normal vector h 1; 1; 0 i,
(iii) it passes through the point P (1; 0; 0), Q(0; 1; 0).
Y-axis
0
-4
4
4
2
-2
2
Normal Vector
Q
P
Z-axis 0
-2
-4
-4
-2
0
X-axis
6. Find the intersection of the planes 3x + y
z
2
4
= 2 and 2x 3y + z = 1.
Answer. Solve the equations for z and equate:
3x + y 2 = z = 2x 3y 1;
Page 3 of 5
i:e:;
3x + y 2 = 2x 3y 1;
Calculus II for Engineering
i:e:;
HOMEWORK 4 – SOLUTION
5x + 4 y = 1 ;
i:e:;
y
Fall, 2010
= 5x4+ 1 :
Putting it into the first equation, we get
2 = 3x + 5x4+ 1
z;
i:e:;
z
= 7x 4 7 :
Using x = t as a parameter, we deduce the line:
x
= t;
y
= 14 54 t;
z
= 47 + 74 t:
= s as a parameter, one can get the line:
1 2
7 + 11 s:
x = + s;
y = s;
z=
5 5
5 5
One can use z = u as a parameter and get the equation of the line.
If one use y
7. Find the distance between the given objects.
(7.1) The point (1; 3; 0) and the plane 3x + y
5z = 2 .
Answer. The distance formula implies
d
(7.2) The planes x + 3y
q + 1(3) 5(0) 2j = p4 :
= j3(1)
35
32 + 12 + ( 5)2
2z = 3 and x + 3y 2z = 1.
Answer. Two planes are parallel. We choose a point P (1; 0; 0) on the plane x
Applying the distance formula to the point P and the other plane, we get
d
+ 3y 2z = 1.
q + 3(0) 2(0) 3j = p2 :
= j1(1)
14
32 + 12 + ( 2)2
8. Find an equation of the plane containing the lines:
: x = 4 + t; y = 2; z = 3 + 2t
M : x = 2 + 2s; y = 2s; z = 1 + 4s:
L
Answer. The vectors parallel to the lines are respectively vL = h 1; 0; 2 i and vM = h 2; 2; 4 i. The
normal vector to the plane is vL vM = h 4; 0; 2 i. We choose one point (4; 2; 3) from L. (One can
choose any point from either line.) Hence, the equation of the plane is
4(x 4) + 0(y 2) + 2(z 3) = 0;
simply;
9. State whether the statement is true or false (not always true).
Page 4 of 5
2x
z
= 5:
Calculus II for Engineering
HOMEWORK 4 – SOLUTION
Fall, 2010
(9.1) Two planes either are parallel or intersect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer. Two planes can be even both parallel and intersect if the planes coincide.
(9.2) The intersection of two planes is a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
F
Answer. It can be a plane if the planes coincide, or can be empty.
(9.3) The intersection of three planes is a point. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. It can be a point, a line, or a plane, or can be empty.
(9.4) Lines that lie in parallel planes are always skew. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T
Answer. It is true, unless the parallel planes coincide.
(9.5) The set of all lines perpendicular to a given line forms a plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. It is false. However, it is true if we take all lines perpendicular to a given line through a
given point.
(9.6) There is one line perpendicular to a given plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
F
Answer. It is false. There is one line perpendicular to a given plane through each point of the
plane.
(9.7) The set of all points equidistant from two given points forms a plane. . . . . . . . . . . . . . . . . . . . . . .
T
10. Determine whether the given lines are the same:
: x = 1 + 4t;
M : x = 9 2s;
L
= 2 2t; z = 2 + 6t
y = 2 + s; z = 8 3s:
y
Answer. The vectors parallel to the lines are vL = h 4; 2; 6 i and vM = h 2; 1; 3 i, respectively.
Since vL = 2vM , the lines L and M are parallel. The point (9; 2; 8) lies on the line M when s = 0.
We solve for t in the x coordinate of the line L to see that 1+4t = 9 implies t = 2. It means when t = 2
and s = 0, the lines L and M have the same x coordinate 9. What about the y and z coordinates when
t = 2 and s = 0? At this time t = 2 and s = 0, the line L has the y and z coordinates, y = 2 2(2) = 2
and z = 2 + 6(2) = 14, respectively. However, the line M has the y and z coordinates, y = 2 and
z = 8. That is,
t
= 2 & s = 0 =)
L
: (x; y; z ) = (9; 2; 14);
It implies that these lines L and M are not the same.
Page 5 of 5
M
: (x; y; z ) = (9; 2; 8):