Download EXAMPLE 6 Find the unit vector in the direction of the vector

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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 6 Find the unit vector in the direction of the vector 2i " j " 2k.
SOLUTION The given vector has length
% 2 i " j " 2k % ! s2
2
! #"1$2 ! #"2$2 ! s9 ! 3
so, by Equation 4, the unit vector with the same direction is
1
3
2
1
2
#2i " j " 2k$ ! 3 i " 3 j " 3 k
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APPLICATIONS
Vectors are useful in many aspects of physics and engineering. In Chapter 13 we will see
how they describe the velocity and acceleration of objects moving in space. Here we look
at forces.
A force is represented by a vector because it has both a magnitude (measured in pounds
or newtons) and a direction. If several forces are acting on an object, the resultant force
experienced by the object is the vector sum of these forces.
50°
32°
T¡
T™
EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the
tensions (forces) T1 and T2 in both wires and their magnitudes.
SOLUTION We first express T1 and T2 in terms of their horizontal and vertical components.
From Figure 20 we see that
100
FIGURE 19
50°
T¡
T™
50°
32°
w
FIGURE 20
32°
% %
% %
! % T % cos 32# i ! % T % sin 32# j
5
T1 ! " T1 cos 50# i ! T1 sin 50# j
6
T2
2
2
.
The resultant T1 ! T2 of the tensions counterbalances the weight w and so we must have
T1 ! T2 ! "w ! 100 j
Thus
("% T1 % cos 50# ! % T2 % cos 32#) i ! (% T1 % sin 50# ! % T2 % sin 32#) j ! 100 j
Equating components, we get
% %
% %
% T % sin 50# ! % T % sin 32# ! 100
Solving the first of these equations for % T % and substituting into the second, we get
T cos 50#
sin 32# ! 100
% T % sin 50# ! % %
" T1 cos 50# ! T2 cos 32# ! 0
1
2
2
1
1
cos 32#
So the magnitudes of the tensions are
%T % !
1
and
100
' 85.64 lb
sin 50# ! tan 32# cos 50#
T cos 50#
% T % ! % cos% 32# ' 64.91 lb
1
2
Substituting these values in (5) and (6), we obtain the tension vectors
T1 ' "55.05 i ! 65.60 j
T2 ' 55.05 i ! 34.40 j
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