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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
a = (26, 10, 4) = (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√hose magnitude is |A − B| = [(−ax − ay + 5az ) · (−ax − ay + 5az )]1/2
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√
= 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
a = (1.5ax + 2.5ay + 0.5az)
+ + )/2.96
= (1.5a
2.5a 0.5a
mp p x y z
(1.5)2 + (2.5)2 +
(0.5)2
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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 = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 31
|(6− 3 B)a−x (2− 3 B)a−y (4 + 3 B)a
| z = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B2 − 8B − 44 = 0. Thus B = 8± 64−176
2 = 11.75 (taking positive option) and so
B 2 2 1
= 3 (11.75)ax − 3 (11.75)ay + 3 (11.75)az = 7.83ax − 7.83ay + 3.92az
1.4. 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 t√y = −aφ · ay = − cos φ. φ
At the point (−
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= 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay). 3,
1),
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 =G = ( —0 .26, 0. 39
, . 0 88)
|(−48, 72, 162)|
c) a unit vector directed from Q
toward P :
(3 1 4)
a QP= P − Q = ,− ,
√ = (0.59, 0.20, −0.78)
26
|P − Q|
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 = 2ax + ay +− 3az and B = ax
3ay + 2az by using the definition of:
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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
a = (26, 10, 4) = (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√hose magnitude is |A − B| = [(−ax − ay + 5az ) · (−ax − ay + 5az )]1/2
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√
= 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
a = (1.5ax + 2.5ay + 0.5az)
+ + )/2.96
= (1.5a
2.5a 0.5a
mp p x y z
(1.5)2 + (2.5)2 +
(0.5)2
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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 = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or
2 2 31
|(6− 3 B)a−x (2− 3 B)a−y (4 + 3 B)a
| z = 10
Expanding, obtain
36 − 8B + 4 B2 + 4 − 8 B + 4 B2 + 16 + 8 B + 1 B2 = 100
9 3 9 √ 3 9
or B2 − 8B − 44 = 0. Thus B = 8± 64−176
2 = 11.75 (taking positive option) and so
B 2 2 1
= 3 (11.75)ax − 3 (11.75)ay + 3 (11.75)az = 7.83ax − 7.83ay + 3.92az
1.4. 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 t√y = −aφ · ay = − cos φ. φ
At the point (−
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= 150◦, and so t = sin 150◦ax − cos 150◦ay = 0.5(ax + 3ay). 3,
1),
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 =G = ( —0 .26, 0. 39
, . 0 88)
|(−48, 72, 162)|
c) a unit vector directed from Q
toward P :
(3 1 4)
a QP= P − Q = ,− ,
√ = (0.59, 0.20, −0.78)
26
|P − Q|
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 = 2ax + ay +− 3az and B = ax
3ay + 2az by using the definition of:
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