Calculus- Early Transcendentals, 2021a

12.3. The Dot Product 429
Solution. We compute
cosθ =(3 · 1 + 3 · 0 + 0 · 0)/( √ 9 + 9 + 0 √ 1 + 0 + 0)
= 3/ √ 18 = 1/ √ 2
so θ = π/4.
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The following are some special cases worth looking at.
Example 12.4
Find the angles between:
1. v and v
2. v and −v
3. v and 0 = 〈0,0,0〉
Solution.
√
1. cosθ = v · v/(|v||v|) =(v 2 1 + v2 2 + v2 3
√v )/( 2 1 + v2 2 + v2 3
v 2 1 + v2 2 + v2 3
)=1, so the angle between v
and itself is zero, which of course is correct.
√
2. cosθ = v·−v/(|v||−v|)=(−v 2 1 −v2 2 −v2 3
√v )/( 2 1 + v2 2 + v2 3
v 2 1 + v2 2 + v2 3
)=−1, so the angle is π,
that is, the vectors point in opposite directions, as of course we already knew.
√ √
3. cosθ = v · 0/(|v||0|) =(0 + 0 + 0)/( v 2 1 + v2 2 + v2 3
0 2 + 0 2 + 0 2 )= 0 , which is undefined. On
0
the other hand, note that since v · 0 = 0 it looks at first as if cosθ will be zero, which as we have
seen means that vectors are perpendicular; only when we notice that the denominator is also zero
do we run into trouble. One way to “fix” this is to adopt the convention that the zero vector 0 is
perpendicular to all vectors; then we can say in general that if v · w = 0, v and w are perpendicular.
Generalizing the examples, note the following useful facts.
•Ifv is parallel or anti-parallel to w then v · w/(|v||w|)=±1, and conversely, if v · w/(|v||w|)=1, v
and w are parallel, while if v · w/(|v||w|)=−1, v and w are anti-parallel.
•Ifv is perpendicular to w then v · w/(|v||w|)=0, and conversely if v · w/(|v||w|)=0thenv and w
are perpendicular.
Given two vectors, it is often useful to find the projection or vector projection of one vector onto
the other, because this turns out to have important meaning in many circumstances. More precisely, given
v and w, we seek a vector parallel to w butwithlengthdeterminedbyv in a natural way, as shown in
Figure 12.5. p is chosen so that the triangle formed by v, p, andv − p is a right triangle.
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