Chapter 3
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1. The x and the y components of a vector a lying on the xy plane are given by
ax  a cos  ,
a y  a sin 
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where a  | a | is the magnitude and  is the angle between a and the positive x axis.
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(a) The x component of a is given by ax  a cos   (7.3 m) cos 250    2.50 m .
(b) Similarly, the y component is given by
a y  a sin   (7.3 m) sin 250    6.86 m   6.9 m.
The results are depicted in the figure below:
In considering the variety of ways to compute these, we note that the vector is 70° below
the – x axis, so the components could also have been found from
ax  (7.3 m) cos 70    2.50 m, a y  (7.3 m) sin 70    6.86 m.
Similarly, we note that the vector is 20° to the left from the – y axis, so one could also
achieve the same results by using
ax  (7.3 m) sin 20    2.50 m, a y  (7.3 m) cos 20    6.86 m.
As a consistency check, we note that
ax2  a y2  ( 2.50 m)2  ( 6.86 m) 2  7.3 m
and
tan  1  a y / ax   tan  1[(  6.86 m) /(  2.50 m)]  250  ,
which are indeed the values given in the problem statement.
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2. (a) With r = 15 m and  = 30°, the x component of r is given by
rx = rcos = (15 m) cos 30° = 13 m.
(b) Similarly, the y component is given by ry = r sin = (15 m) sin 30° = 7.5 m.
3. A vector a can be represented in the magnitude-angle notation (a, ), where
a  ax2  a y2
is the magnitude and
 ay 

 ax 
  tan 1 
is the angle a makes with the positive x axis.
(a) Given Ax = 25.0 m and Ay = 40.0 m, A  ( 25.0 m) 2  (40.0 m) 2  47.2 m.
(b) Recalling that tan  = tan ( + 180°),
tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°.
Noting that the vector is in the third quadrant (by the signs of its x and y components) we
see that 122° is the correct answer. The graphical calculator “shortcuts” mentioned above
are designed to correctly choose the right possibility. The results are depicted in the
figure below:
We can check our answers by noting that the x- and the y- components of A can be
written as
Ax  A cos  ,
Ay  A sin 
Substituting the results calculated above, we obtain
Ax  (47.2 m) cos122    25.0 m, Ay  (47.2 m) sin122    40.0 m
which indeed are the values given in the problem statement.
6. (a) The height is h = d sin, where d = 12.5 m and  = 20.0°. Therefore, h = 4.28 m.
(b) The horizontal distance is d cos = 11.7 m.
9. All distances in this solution are understood to be in meters.
ˆ m.
(a) a  b  [4.0  (1.0)] ˆi  [(3.0)  1.0] ˆj  (1.0  4.0)kˆ  (3.0iˆ  2.0ˆj  5.0 k)
ˆ m.
(b) a  b  [4.0  (1.0)]iˆ  [(3.0) 1.0]jˆ  (1.0  4.0)kˆ  (5.0 iˆ  4.0 ˆj  3.0 k)
  
  
(c) The requirement a  b  c  0 leads to c  b  a , which we note is the opposite of
ˆ m.
what we found in part (b). Thus, c  (5.0 ˆi  4.0 ˆj  3.0 k)
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11. We write r  a  b . When not explicitly displayed, the units here are assumed to be
meters.
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(a) The x and the y components of r are rx = ax + bx = (4.0 m) – (13 m) = –9.0 m and ry =
ay + by = (3.0 m) + (7.0 m) = 10 m, respectively. Thus r  ( 9.0 m) ˆi  (10 m) ˆj .
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(b) The magnitude of r is
r | r | rx2  ry2  (9.0 m)2  (10 m)2  13 m .
(c) The angle between the resultant and the +x axis is given by
 ry
 rx
  tan 1 

 1  10.0 m 
  tan 
   48  or 132 .
  9.0 m 

Since the x component of the resultant is negative and the y component is positive,
characteristic of the second quadrant, we find the angle is 132° (measured
counterclockwise from +x axis).
The addition of the two vectors is depicted in the figure below (not to scale). Indeed, we
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expect r to be in the second quadrant.
16. (a) a  b  (3.0 ˆi  4.0 ˆj) m  (5.0 ˆi  2.0 ˆj) m  (8.0 m) ˆi  (2.0 m) ˆj.
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(b) The magnitude of a  b is
| a  b | (8.0 m) 2  (2.0 m) 2  8.2 m.
(c) The angle between this vector and the +x axis is
tan–1[(2.0 m)/(8.0 m)] = 14°.
(d) b  a  (5.0 ˆi  2.0 ˆj) m  (3.0 ˆi  4.0 ˆj) m  (2.0 m) ˆi  (6.0 m)ˆj .
(e) The magnitude of the difference vector b  a is
| b  a | (2.0 m) 2  (6.0 m) 2  6.3 m.
(f) The angle between this vector and the +x axis is tan-1[( –6.0 m)/(2.0 m)] = –72°. The
vector is 72° clockwise from the axis defined by î .
46. The vectors are shown on the diagram. The x axis runs from west to east and the y
axis runs from south to north. Then ax = 5.0 m, ay = 0,
bx = –(4.0 m) sin 35° = –2.29 m, by = (4.0 m) cos 35° = 3.28 m.
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(a) Let c  a  b . Then cx  ax  bx = 5.00 m  2.29 m = 2.71 m and
c y  a y  by = 0 + 3.28 m = 3.28 m . The magnitude of c is
c  cx2  cy2 
 2.71m
2
  3.28m   4.2 m.
2
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(b) The angle  that c  a  b makes with the +x axis is
 cy
 cx
  tan 1 

1  3.28 
  tan 
  50.5  50.
 2.71 

The second possibility ( = 50.4° + 180° = 230.4°) is rejected because it would point in a
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direction opposite to c .
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(c) The vector b  a is found by adding a to b . The result is shown on the diagram to
the right. Let c  b  a. The components are
cx  bx  ax  2.29 m  5.00 m  7.29 m
c y  by  a y  3.28 m.
The magnitude of c is c  cx2  c 2y  8.0 m .
(d) The tangent of the angle  that c makes with the +x axis (east) is
tan  
cy
cx

3.28 m
 4.50.
7.29 m
There are two solutions: –24.2°
and 155.8°. As the diagram shows, the second solution is
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correct. The vector c  a  b is 24° north of west.