James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 12.2 Vectors 801
The magnitude or length of the vector v is the length of any of its representations and
is denoted by the symbol | v | or i v i. By using the distance formula to compute the length
of a segment OP, we obtain the following formulas.
The length of the two-dimensional vector a − k a 1 , a 2 l is
y
(a¡+b¡, a+b)
| a | − sa 2 1 1 a 2 2
The length of the three-dimensional vector a − ka 1 , a 2 , a 3 l is
a+b
b
b
| a | − sa 2 1
1 a 2 2
1 a 2 3
a
a
0 a¡
FIGURE 14
b¡
b¡
a
x
How do we add vectors algebraically? Figure 14 shows that if a − ka 1 , a 2 l and
b − kb 1 , b 2 l, then the sum is a 1 b − ka 1 1 b 1 , a 2 1 b 2 l, at least for the case where the
components are positive. In other words, to add algebraic vectors we add corresponding
components. Similarly, to subtract vectors we subtract corresponding components. From
the similar triangles in Figure 15 we see that the components of ca are ca 1 and ca 2 . So to
multiply a vector by a scalar we multiply each component by that scalar.
If a − k a 1 , a 2 l and b − kb 1 , b 2 l, then
a a
ca
ca
a 1 b − ka 1 1 b 1 , a 2 1 b 2 l a 2 b − ka 1 2 b 1 , a 2 2 b 2 l
ca − k ca 1 , ca 2 l
a¡
FIGURE 15
ca¡
Similarly, for three-dimensional vectors,
k a 1 , a 2 , a 3 l 1 k b 1 , b 2 , b 3 l − k a 1 1 b 1 , a 2 1 b 2 , a 3 1 b 3 l
k a 1 , a 2 , a 3 l 2 k b 1 , b 2 , b 3 l − k a 1 2 b 1 , a 2 2 b 2 , a 3 2 b 3 l
ck a 1 , a 2 , a 3 l − k ca 1 , ca 2 , ca 3 l
Example 4 If a − k 4, 0, 3 l and b − k 22, 1, 5 l, find | a | and the vectors a 1 b,
a 2 b, 3b, and 2a 1 5b.
SOLUTION | a | − s4 2 1 0 2 1 3 2 − s25 − 5
a 1 b − k4, 0, 3l 1 k22, 1, 5l
− k4 1 s22d, 0 1 1, 3 1 5l − k2, 1, 8l
a 2 b − k4, 0, 3l 2 k22, 1, 5l
− k4 2 s22d, 0 2 1, 3 2 5l − k6, 21, 22l
3b − 3k22, 1, 5l − k3s22d, 3s1d, 3s5dl − k26, 3, 15l
2a 1 5b − 2k4, 0, 3l 1 5k22, 1, 5l
− k8, 0, 6l 1 k210, 5, 25l − k22, 5, 31l ■
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