College Trigonometry, 2011a

1038 Applications of Trigonometry
and Q(1,m 1 + b 1 ). We let ⃗v 1 = −→ PQ = 〈1 − 0, (m1 + b 1 ) − b 1 〉 = 〈1,m 1 〉, and note that since ⃗v 1 is
determined by two points on L 1 , it may be viewed as lying on L 1 . Hence it has the same direction
as L 1 . Similarly, we get the vector ⃗v 2 = 〈1,m 2 〉 which has the same direction as the line L 2 . Hence,
L 1 and L 2 are perpendicular if and only if ⃗v 1 ⊥ ⃗v 2 . According to Theorem 11.25, ⃗v 1 ⊥ ⃗v 2 if and
only if ⃗v 1 · ⃗v 2 = 0. Notice that ⃗v 1 · ⃗v 2 = 〈1,m 1 〉·〈1,m 2 〉 =1+m 1 m 2 . Hence, ⃗v 1 · ⃗v 2 = 0 if and only
if 1 + m 1 m 2 = 0, which is true if and only if m 1 m 2 = −1, as required.
While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically,
there is more to the story of the dot product. Consider the two nonzero vectors ⃗v and ⃗w drawn
with a common initial point O below. For the moment, assume that the angle between ⃗v and ⃗w,
which we’ll denote θ, is acute. We wish to develop a formula for the vector ⃗p, indicated below,
which is called the orthogonal projection of ⃗v onto ⃗w. The vector ⃗p is obtained geometrically
as follows: drop a perpendicular from the terminal point T of ⃗v to the vector ⃗w and call the point
of intersection R. The vector ⃗p is then defined as ⃗p = −→ OR. Like any vector, ⃗p is determined by its
magnitude ‖⃗p‖ and its direction ˆp accordingtotheformula⃗p = ‖⃗p‖ˆp. Since we want ˆp to have the
same direction as ⃗w, wehaveˆp = ŵ. To determine ‖⃗p‖, we make use of Theorem 10.4 as applied
to the right triangle △ORT. We find cos(θ) = ‖⃗p‖
‖⃗v‖
,or‖⃗p‖ = ‖⃗v‖ cos(θ). To get things in terms of
‖⃗v‖‖ ⃗w‖ cos(θ) ⃗v· ⃗w
just ⃗v and ⃗w, we use Theorem 11.23 to get ‖⃗p‖ = ‖⃗v‖ cos(θ) =
‖ ⃗w‖
=
‖ ⃗w‖
. Using Theorem
( )
⃗v· ⃗w
11.22, werewrite
‖ ⃗w‖ = ⃗v · 1
‖ ⃗w‖ ⃗w = ⃗v · ŵ. Hence, ‖⃗p‖ = ⃗v · ŵ, and since ˆp = ŵ, wenowhavea
formula for ⃗p completely in terms of ⃗v and ⃗w, namely ⃗p = ‖⃗p‖ˆp =(⃗v · ŵ)ŵ.
⃗v
⃗v
T
T
⃗w
⃗w
‖⃗v‖
O
θ
O
θ
⃗p = −→ OR
R
O
θ
‖⃗p‖
R
Now suppose that the angle θ between ⃗v and ⃗w is obtuse, and consider the diagram below. In this
case, we see that ˆp = −ŵ and using the triangle △ORT, we find ‖⃗p‖ = ‖⃗v‖ cos(θ ′ ). Since θ+θ ′ = π,
it follows that cos(θ ′ )=− cos(θ), which means ‖⃗p‖ = ‖⃗v‖ cos(θ ′ )=−‖⃗v‖ cos(θ). Rewriting this
last equation in terms of ⃗v and ⃗w as before, we get ‖⃗p‖ = −(⃗v · ŵ). Putting this together with
ˆp = −ŵ, weget⃗p = ‖⃗p‖ˆp = −(⃗v · ŵ)(−ŵ) =(⃗v · ŵ)ŵ in this case as well.