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

808 Chapter 12 Vectors and the Geometry of Space
These properties are easily proved using Definition 1. For instance, here are the proofs
of Properties 1 and 3:
1. a ? a − a 2 1 1 a 2 2 1 a 2 3 − | a | 2
3. a ? sb 1 cd − ka 1 , a 2 , a 3 l ? kb 1 1 c 1 , b 2 1 c 2 , b 3 1 c 3 l
− a 1 sb 1 1 c 1 d 1 a 2 sb 2 1 c 2 d 1 a 3 sb 3 1 c 3 d
− a 1 b 1 1 a 1 c 1 1 a 2 b 2 1 a 2 c 2 1 a 3 b 3 1 a 3 c 3
− sa 1 b 1 1 a 2 b 2 1 a 3 b 3 d 1 sa 1 c 1 1 a 2 c 2 1 a 3 c 3 d
− a ? b 1 a ? c
The proofs of the remaining properties are left as exercises.
■
x
z
b
¨
O
B
a
a-b
A
y
The dot product a ? b can be given a geometric interpretation in terms of the angle
between a and b, which is defined to be the angle between the representations of a and
b that start at the origin, where 0 < < . In other words, is the angle between the
line segments OA l and OB l in Figure 1. Note that if a and b are parallel vectors, then
− 0 or − .
The formula in the following theorem is used by physicists as the definition of the dot
product.
FIGURE 1
3 Theorem If is the angle between the vectors a and b, then
a ? b − | a | | b | cos
Proof If we apply the Law of Cosines to triangle OAB in Figure 1, we get
4 | AB | 2 − | OA | 2 1 | OB | 2 2 2 | OA | | OB | cos
(Observe that the Law of Cosines still applies in the limiting cases when − 0 or , or
a − 0 or b − 0.) But | OA | − | a | , | OB | − | b | , and | AB | − | a 2 b | , so Equation 4
becomes
5 | a 2 b | 2 − | a | 2 1 | b | 2 2 2 | a | | b | cos
Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this
equation as follows:
| a 2 b | 2 − sa 2 bd ? sa 2 bd
− a ? a 2 a ? b 2 b ? a 1 b ? b
Therefore Equation 5 gives
− | a | 2 2 2a ? b 1 | b | 2
Thus
| a | 2 2 2a ? b 1 | b | 2 − | a | 2 1 | b | 2 2 2 | a | | b | cos
22a ? b − 22 | a | | b | cos
or a ? b − | a | | b | cos ■
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