p03_019.pdf

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Chapter 3 - 3.19
19. Many of the operations are done eciently on most modern graphical calculators using their built-in
vector manipulation and rectangular
↔
(a) Using unit-vector notation,
a = 50 cos (30 ◦ )i + 50 sin (30 ◦ )j
b = 50 cos (195 ◦ )i + 50 sin (195 ◦ )j
c = 50 cos (315 ◦ )i + 50 sin (315 ◦ )j
a + b + c =30 . 4 i
−
23 . 3 j .
23 . 3) 2 =38m.
(b) The two possibilities presented by a simple calculation for the angle between the vector described
in part (a) and the + x direction are tan − 1 (
−
37 . 5 ◦ ) = 142 . 5 ◦ .
The former possibility is the correct answer since the vector is in the fourth quadrant (indicated
by the signs of its components). Thus, the angle is
−
23 . 2 / 30 . 4) =
−
37 . 5 ◦ , and 180 ◦ +(
−
37 . 5 ◦ , which is to say that it is roughly 38 ◦
clockwise from the + x axis. This is equivalent to 322 . 5 ◦ counterclockwise from + x .
(c) We find a
−
notation. The magnitude of this result is √ 127 2 +2 . 6 2
b + c =(43 . 3
−
(
−
48 . 3) + 35 . 4)i
−
(25
−
(
−
12 . 9) + (
−
35 . 4))j = 127i+2 . 6j in unit-vector
10 2 m.
(d) The angle between the vector described in part (c) and the + x axis is tan − 1 (2 . 6 / 127)
≈
1 . 3
×
≈
1 ◦ .
(e) Using unit-vector notation, d is given by
d = a + b
−
c
=
−
40 . 4i+47 . 4j
40 . 4) 2 +47 . 4 2 =62m.
(f) The two possibilities presented by a simple calculation for the angle between the vector described
in part (e) and the + x axis are tan − 1 (47 . 4 / (
−
50 ◦ ) = 130 ◦ .Wechoose
the latter possibility as the correct one since it indicates that d is in the second quadrant (indicated
by the signs of its components).
−
40 . 4)) =
−
50 ◦ , and 180 ◦ +(
−
polar “shortcuts.” In this solution, we employ the “traditional”
methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood.
The magnitude of this result is 30 . 4 2 +(
−
which has a magnitude of (
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