Chapter 3: Vectors
Example Questions & Problems
a x  a cos 
a y  a sin 
a  a 2x  a 2y
tan  
ay
ax
a  a x ˆi  a x ˆj  a xkˆ
Example 3.1
The two vectors shown lie in the xy plane. What are the signs of the x
and y components of (a) d1  d2 , (b) d1  d2 , and (c) d2  d1 ?
Example 3.2
A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2
km south. (a) Sketch the vector diagram that represents this motion. (b) How far and (c)
in what direction would a bird fly in a straight line from the same starting point to the
same final point?
Example 3.3
A particle undergoes three successive displacements in a plane, as follows: d1 (4.00 m
southwest), then d2 (5.00 m east), and finally d3 (6.00 m in a direction 60o north of east).
(a) Sketch the vector diagram that represents this motion and (b) the determine the
direction and magnitude of the particle’s net displacement?
3-1
Example 3.4
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground
cannot be distinguished from the sky) while returning to base camp. He was supposed to
travel due north for 5.6 km, but when the snow clears; he discovers that he actually
traveled 7.8 km at 50o north of due east. (Sketch the vector diagram that represents this
motion.) (a) How far and (b) in what direction must he now travel to reach base camp?
Solution
According to the vector diagram, the goal was to move along the vector A , but the
whiteout caused the explorer to move along the vector B . To reach the base camp
located at A, the third vector C must be traveled such that
A BC 
C  A B
The magnitude and directions (angles) of each vector is
magnitude
angle
A = 5.6 km
A = 900
A
B = 7.8 km
B = 500
B
The components of the two vectors are
A
B
x-component
A cos A = 5.6 cos 900 = 0 km
B cos B = 7.8 cos 500 = 5.01 km
y-component
A sin A = 5.6 sin 900 = 5.6 km
B sin B = 7.8 sin 500 = 5.98 km
To find the magnitude and direction C (subtract the column values), we compute
C  A  B  (A x  B x , A y  B y )
 (0  5.01, 5.6  5.98)  (  5.01,  0.38) km
The magnitude and direction of the vector C is
C
C2x  C2y 
 5.01
2
  0.38   5.0 k m  C
2
and the direction is
  tan-1
Cy
Cx
 0.38 
  180  4.3  180  184  θ
 5.01 
 tan-1 
3-2