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 ...
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 + 3as) − (2ax + 3ay − 2as) = (−ax − ay + 5as)
1/2
,
whose magnitude is |A − B| = (−a − a + 5a ) ∙ (−a – a + 5a ) = 1 + 1 + 25 =
, x y s x y s
3 3 = 5.20. The unit vector is therefore
aAB = (−ax − ay + 5as)/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 + 0.5as) = (1.5a + 2.5a + 0.5a )/2.96
a = , x y s
mp
(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 − 2 B)ax − (2 − 2 B)ay − (4 + 1 B)as| = 10
3 3 3
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 = 11.75 (taking positive option) and so
2
2 2 1
B = (11.75)a − (11.75)a + (11.75)a = 7.83a – 7.83a + 3.92a
x y s x y s
3 3 3
2
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