CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitude of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
√
w√h o s e magnitude is |A − B| = [(−ax − ay + 5az) · (−ax − ay + 5az)]1/2 = 1 + 1 + 25 =
3 3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
(1.5ax + 2.5ay + 0.5az )
a = = (1.5a + 2.5a + 0.5a )/2.96
mp p x y z
(1.5)2 + (2.5)2 + (0.5)2
1.3. The vector from the origin to the point A is given as (6,− 2,−4), and the unit vector directed from
the origin toward point B is (2,— 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, −2, −4) and B = 13B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 1
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az | = 10
Expanding, 4obtain
36 − 8B + B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B − 8B − 44 = 0. Thus B =
2
2
8± 64−176
= 11.75 (taking positive option) and so
2 2 1
B
= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
1
,1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine √the unit
vector in rectangular components that lies in the xy plane, is tangent to the circle at (— 3, 1, 0),
and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = −aφ. Its√x and
y components are tx = −aφ · ax = sin φ, and ty = −aφ · ay = − cos φ. At the point (− 3, 1),
√
φ = 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P (1, 2, −1)
and Q(−2, 1, 3), find:
a) G at P : G(1, 2, −1) = (48, 36, 18)
b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so
(−48, 72, 162)
a = = ( 0 26 0 39 0 88)
G —. , . , .
|(−48, 72, 162)|
c) a unit vector directed from Q toward P :
P−Q (3, −1, 4)
aQP = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acute angle between the two vectors A = 2a x + ay + 3az and B = ax — 3ay + 2az by using
the definition of:
√ √
a) the dot pro duct √ 2 − 3 + 6 = 5 = AB cos θ, where A = 22 + 12 + 32 = 14,
: First, A · B = √
and where B = 12 + 32 + 22 = 14. Therefore cos θ = 5/14, so that θ = 69.1◦.
b) the cross product: Begin with
A Ø ax ay az Ø
B a a 7a
× 2 = 1 3 = 11 x − y − z
Ø 1 −3 2 Ø
√ √ √
and then |A × ° B√| = 11
¢ 2 + 12 ◦+ 72 = 171. So now, with |A × B| = AB sin θ = 171,
find θ = sin−1 171/14 = 69.1
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region | x| , |y |, and |z |
less than 2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
2
, 1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = − 3ax 2ay + 4az and B = 2ax + a−y 2az. Does this
ambiguity exist when the dot product is used?
We use the relation A × B = |A||B| sin θn. With the given vectors we find
∑ ∏
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
| {z }
± n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really w a √
n t t h √e mag
√ nitude of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
√
In using the d√o t product, we find A · B = 6 − 2 − 8 = −4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
Find:
a) a unit vector in the direction of G at P (3, 4, −2): Have Gp = 25/(9 + 16) ×(3, 4, 0) = 3ax + 4ay,
and |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is found through aG · ax = cos θ. So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦.
c) the value of the following double integral on the plane y = 7:
Z 4 Z 2
G · aydzdx
0 0
Z 4 Z 2 Z 4Z 2 Z 4
25 a ) a 25 350
(xa + y · dzdx = × 7 dzdx = dx
0 x + 49 0 x + 49
x y y
0 0 x2 + y2 0
2 2
∑ µ ∂ ∏
1 −1 4
= 350 × tan − 0 = 26
7 7
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = √b ( a x + √ ay + az) and B =
b(ax − ay + az). Now use A · B = |A||B| cos θ, or b2(1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
1/3 ⇒ θ = 70.53◦. This result (in magnitude) is the same for any two diagonal vectors.
3
, 1.11. Given the points M (0.1, −0.2, −0.1), N (−0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find:
a) the vector RMN : RMN = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
b) the dot product RMN · RMP : RMP = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). RMN ·
RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP :
(0.3, 0.2, 0.2) 0.05
RMN · aRMP = (−0.3, 0.3, 0.4) · √ = √ = 0.12
0.09 + 0.04 + 0.04 0.17
d) the angle between RMN and RMP :
µ ∂ µ ∂
−1 RMN · RMP −1 0.05
θ √ = 78◦
M = cos = cos √
|R MN ||R MP | 0.34 0.17
1.12. Write an expression in rectangular components for the vector that extends from (x1, y1, z1) to
(x2, y2, z2) and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A1 = x1ax + y1ay + z1az and A2 = x2ax + y2ay + z2az
The desired vector will now be the difference:
A12 = A2 − A1 = (x2 − x1)ax + (y2 − y1)ay + (z2 − z1)az
whose magnitude is
p £ §1/2
|A 12 | = A 12 · A 12 = (x2 − x1)2 + (y2 − y1 )2 + (z 2 − z 1)2
1.13. a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3):
F·G (10, −6, 5) · (0.1, 0.2, 0.3)
F = G= (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
||G
|G|2 0.01 + 0.04 + 0.09
b) Find the vector component of F that is perpendicular to G:
FpG = F − F G|| = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
G =G−G = G − G · F F = (0.1, 0.2, 0.3) − 1.3 (10, −6, 5) = (0.02, 0.25, 0.26)
pF ||F
|F|2 100 + 36 + 25
4
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitude of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A − B): First
A − B = (ax + 2ay + 3az) − (2ax + 3ay − 2az) = (−ax − ay + 5az)
√
w√h o s e magnitude is |A − B| = [(−ax − ay + 5az) · (−ax − ay + 5az)]1/2 = 1 + 1 + 25 =
3 3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5az)/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
(1.5ax + 2.5ay + 0.5az )
a = = (1.5a + 2.5a + 0.5a )/2.96
mp p x y z
(1.5)2 + (2.5)2 + (0.5)2
1.3. The vector from the origin to the point A is given as (6,− 2,−4), and the unit vector directed from
the origin toward point B is (2,— 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, −2, −4) and B = 13B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 1
|(6 − 3 B)ax − (2 − 3 B)ay − (4 + 3 B)az | = 10
Expanding, 4obtain
36 − 8B + B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B − 8B − 44 = 0. Thus B =
2
2
8± 64−176
= 11.75 (taking positive option) and so
2 2 1
B
= (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az
3 3 3
1
,1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine √the unit
vector in rectangular components that lies in the xy plane, is tangent to the circle at (— 3, 1, 0),
and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = −aφ. Its√x and
y components are tx = −aφ · ax = sin φ, and ty = −aφ · ay = − cos φ. At the point (− 3, 1),
√
φ = 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az. Given two points, P (1, 2, −1)
and Q(−2, 1, 3), find:
a) G at P : G(1, 2, −1) = (48, 36, 18)
b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so
(−48, 72, 162)
a = = ( 0 26 0 39 0 88)
G —. , . , .
|(−48, 72, 162)|
c) a unit vector directed from Q toward P :
P−Q (3, −1, 4)
aQP = = √ = (0.59, 0.20, −0.78)
|P − Q| 26
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acute angle between the two vectors A = 2a x + ay + 3az and B = ax — 3ay + 2az by using
the definition of:
√ √
a) the dot pro duct √ 2 − 3 + 6 = 5 = AB cos θ, where A = 22 + 12 + 32 = 14,
: First, A · B = √
and where B = 12 + 32 + 22 = 14. Therefore cos θ = 5/14, so that θ = 69.1◦.
b) the cross product: Begin with
A Ø ax ay az Ø
B a a 7a
× 2 = 1 3 = 11 x − y − z
Ø 1 −3 2 Ø
√ √ √
and then |A × ° B√| = 11
¢ 2 + 12 ◦+ 72 = 171. So now, with |A × B| = AB sin θ = 171,
find θ = sin−1 171/14 = 69.1
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region | x| , |y |, and |z |
less than 2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
2
, 1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = − 3ax 2ay + 4az and B = 2ax + a−y 2az. Does this
ambiguity exist when the dot product is used?
We use the relation A × B = |A||B| sin θn. With the given vectors we find
∑ ∏
√ 2ay + az √ √
A × B = 14ay + 7az = 7 5 √ = 9 + 4 + 16 4 + 1 + 4 sin θ n
5
| {z }
± n
where n is identified as shown; we see that n can be positive or negative, as sin θ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really w a √
n t t h √e mag
√ nitude of the angle anyway. Choosing the positive sign, we are left with
sin θ = 7 5/( 29 9) = 0.969. Two values of θ (75.7◦ and 104.3◦) satisfy this equation, and
hence the real ambiguity.
√
In using the d√o t product, we find A · B = 6 − 2 − 8 = −4 = |A||B| cos θ = 3 29 cos θ, or
cos θ = −4/(3 29) = −0.248 ⇒ θ = −75.7◦. Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one θ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
25
G= (xax + yay)
(x2 + y2)
Find:
a) a unit vector in the direction of G at P (3, 4, −2): Have Gp = 25/(9 + 16) ×(3, 4, 0) = 3ax + 4ay,
and |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is found through aG · ax = cos θ. So cos θ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus θ = 53◦.
c) the value of the following double integral on the plane y = 7:
Z 4 Z 2
G · aydzdx
0 0
Z 4 Z 2 Z 4Z 2 Z 4
25 a ) a 25 350
(xa + y · dzdx = × 7 dzdx = dx
0 x + 49 0 x + 49
x y y
0 0 x2 + y2 0
2 2
∑ µ ∂ ∏
1 −1 4
= 350 × tan − 0 = 26
7 7
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = √b ( a x + √ ay + az) and B =
b(ax − ay + az). Now use A · B = |A||B| cos θ, or b2(1 − 1 + 1) = ( 3b)( 3b) cos θ ⇒ cos θ =
1/3 ⇒ θ = 70.53◦. This result (in magnitude) is the same for any two diagonal vectors.
3
, 1.11. Given the points M (0.1, −0.2, −0.1), N (−0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find:
a) the vector RMN : RMN = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4).
b) the dot product RMN · RMP : RMP = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). RMN ·
RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP :
(0.3, 0.2, 0.2) 0.05
RMN · aRMP = (−0.3, 0.3, 0.4) · √ = √ = 0.12
0.09 + 0.04 + 0.04 0.17
d) the angle between RMN and RMP :
µ ∂ µ ∂
−1 RMN · RMP −1 0.05
θ √ = 78◦
M = cos = cos √
|R MN ||R MP | 0.34 0.17
1.12. Write an expression in rectangular components for the vector that extends from (x1, y1, z1) to
(x2, y2, z2) and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A1 = x1ax + y1ay + z1az and A2 = x2ax + y2ay + z2az
The desired vector will now be the difference:
A12 = A2 − A1 = (x2 − x1)ax + (y2 − y1)ay + (z2 − z1)az
whose magnitude is
p £ §1/2
|A 12 | = A 12 · A 12 = (x2 − x1)2 + (y2 − y1 )2 + (z 2 − z 1)2
1.13. a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3):
F·G (10, −6, 5) · (0.1, 0.2, 0.3)
F = G= (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
||G
|G|2 0.01 + 0.04 + 0.09
b) Find the vector component of F that is perpendicular to G:
FpG = F − F G|| = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
G =G−G = G − G · F F = (0.1, 0.2, 0.3) − 1.3 (10, −6, 5) = (0.02, 0.25, 0.26)
pF ||F
|F|2 100 + 36 + 25
4
