CHAPTER 1 – 9th Edition
1.1. If 𝐀 represents a vector two units in length directed due west, 𝐁 represents a vector three units in length
directed due north, and 𝐀 + 𝐁 = 𝐂 − 𝐃 and 2𝐁 − 𝐀 = 𝐂 + 𝐃, find the magnitudes and directions of
𝐂 and 𝐃. Take north as the positive 𝑦 direction:
With north as positive 𝑦, west will be -𝑥. We may therefore set up:
𝐂 + 𝐃 = 2𝐁 − 𝐀 = 6𝐚𝑦 + 2𝐚𝑥 and
𝐂 − 𝐃 = 𝐀 + 𝐁 = −2𝐚𝑥 + 3𝐚𝑦
Add the equations to find 𝐂 = 4.5𝐚𝑦 (north), and then 𝐃 = 2𝐚𝑥 + 1.5𝐚𝑦 (east of northeast).
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 )
[ ]1∕2 √
whose magnitude is |𝐀 − 𝐁| = (−𝐚𝑥 − 𝐚𝑦 + 5𝐚𝑧 ) ⋅ (−𝐚𝑥 − 𝐚𝑦 + 5𝐚𝑧 ) = 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 )
𝐚𝑚𝑝 = √ = (1.5𝐚𝑥 + 2.5𝐚𝑦 + 0.5𝐚𝑧 )∕2.96
(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 = 13 𝐵(2, −2, 1), we use the fact that |𝐁 − 𝐀| = 10, or
|(6 − 23 𝐵)𝐚𝑥 − (2 − 32 𝐵)𝐚𝑦 − (4 + 13 𝐵)𝐚𝑧 | = 10
Expanding, obtain
36 − 8𝐵 + 94 𝐵 2 + 4 − 38 𝐵 + 49 𝐵 2 + 16 + 83 𝐵 + 19 𝐵 2 = 100
√
or 𝐵 2 − 8𝐵 − 44 = 0. Thus 𝐵 = 8± 64−176
2
= 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
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
, 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 𝑥:
A unit vector tangent to this circle in the general increasing 𝑥 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 = 330◦ , and so 𝐭 = − sin 330◦ 𝐚𝑥 + cos 330◦ 𝐚𝑦 = 0.5(𝐚𝑥 + 3𝐚𝑦 ).
1.5. An equilateral triangle lies in the 𝑥𝑦 plane with its centroid at the origin. One vertex lies on the positive
𝑦 axis.
y
a) Find unit vectors that are directed from the origin to
the three vertices: Referring to the figure,the easy a1
one is 𝐚1 = 𝐚𝑦 . Then, 𝐚2 will have negative 𝑥 and 𝑦
components, and can be constructed as 𝐚2 = 𝐺(−𝐚𝑥 a5 a6
− tan 30◦ 𝐚𝑦 ) where 𝐺 = (1 + tan 30◦ )1∕2 = 0.87.
So finally 𝐚2 = −0.87(𝐚𝑥 + 0.58𝐚𝑦 ). Then, 𝐚3 is the
x
same as 𝐚2 , but with the 𝑥 component reversed:
30°
𝐚3 = 0.87(𝐚𝑥 − 0.58𝐚𝑦 ).
a a3
b) Find unit vectors that are directed from the origin 2
to the three sides, intersecting these at right angles: a4
These will be 𝐚4 , 𝐚5 , and 𝐚6 in the figure, which are in turn just the part 𝑎 results, oppositely
directed:
𝐚4 = −𝐚1 = −𝐚𝑦 , 𝐚5 = −𝐚3 = −0.87(𝐚𝑥 − 0.58𝐚𝑦 ), and 𝐚6 = −𝐚2 = +0.87(𝐚𝑥 + 0.58𝐚𝑦 ).
1.6. Find the acute angle between the two vectors 𝐀 = 2𝐚𝑥 + 𝐚𝑦 + 3𝐚𝑧 and 𝐁 = 𝐚𝑥 − 3𝐚𝑦 + 2𝐚𝑧 by using
the definition of:
√ √
a) the dot product:√First, 𝐀 ⋅ 𝐁 = 2 −
√ 3 + 6 = 5 = 𝐴𝐵 cos 𝜃, where 𝐴 = 2 2 + 1 2 + 32 = 14,
and where 𝐵 = 1 + 3 + 2 = 14. Therefore cos 𝜃 = 5∕14, so that 𝜃 = 69.1 .
2 2 2 ◦
b) the cross product: Begin with
|𝐚 𝐚 𝐚 |
| 𝑥 𝑦 𝑧|
| |
𝐀 × 𝐁 = | 2 1 3 | = 11𝐚𝑥 − 𝐚𝑦 − 7𝐚𝑧
| |
| 1 −3 2 |
| |
√ √ √
and then |𝐀 × 𝐁|
(√= 11 )+ 1 + 7 = 171. So now, with |𝐀 × 𝐁| = 𝐴𝐵 sin 𝜃 = 171,
2 2 2
find 𝜃 = sin−1 171∕14 = 69.1◦
1.7. Given the field 𝐅 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 . If 𝐅 ⋅ 𝐆 = 2𝑥𝑦 and 𝐅 × 𝐆 = (𝑥2 − 𝑦2 ) 𝐚𝑧 , find 𝐆:
Let 𝐆 = 𝑔1 𝐚𝑥 + 𝑔2 𝐚𝑦 + 𝑔3 𝐚𝑧 Then 𝐅 ⋅ 𝐆 = 𝑔1 𝑥 + 𝑔2 𝑦 = 2𝑥𝑦, and
|𝐚 𝐚 𝐚 |
| 𝑥 𝑦 𝑧|
| |
𝐅 × 𝐆 = | 𝑥 𝑦 0 | = 𝑔3 𝑦 𝐚𝑥 − 𝑔3 𝑥 𝐚𝑦 + (𝑔2 𝑥 − 𝑔1 𝑦) 𝐚𝑧 = (𝑥2 − 𝑦2 ) 𝐚𝑧
| |
|𝑔1 𝑔2 𝑔3 |
| |
From the last equation, it is clear that 𝑔3 = 0, and that 𝑔1 = 𝑦 and 𝑔2 = 𝑥. This is confirmed in the
𝐅 ⋅ 𝐆 equation. So finally 𝐆 = 𝑦𝐚𝑥 + 𝑥𝐚𝑦 .
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