Michael Corral: Vector Calculus

16 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Theorem1.6. Letv,wbenonzerovectors,andletθbetheanglebetweenthem. Then
cosθ= v·w
‖v‖‖w‖
(1.8)
Proof: We will prove the theorem for vectors in 3 (the proof for 2 is similar). Let
v=(v 1 ,v 2 ,v 3 ) and w=(w 1 ,w 2 ,w 3 ). By the Law of Cosines (see Figure 1.3.2), we have
‖v−w‖ 2 =‖v‖ 2 +‖w‖ 2 −2‖v‖‖w‖cosθ (1.9)
(note that equation (1.9) holds even for the “degenerate” casesθ=0 ◦ and 180 ◦ ).
z
0
v
θ
w
v−w
y
x
Figure 1.3.2
Since v−w=(v 1 −w 1 ,v 2 −w 2 ,v 3 −w 3 ), expanding‖v−w‖ 2 in equation (1.9) gives
‖v‖ 2 +‖w‖ 2 −2‖v‖‖w‖cosθ=(v 1 −w 1 ) 2 +(v 2 −w 2 ) 2 +(v 3 −w 3 ) 2
= (v 2 1−2v 1 w 1 +w 2 1)+(v 2 2−2v 2 w 2 +w 2 2)+(v 2 3−2v 3 w 3 +w 2 3)
= (v 2 1+v 2 2+v 2 3)+(w 2 1+w 2 2+w 2 3)−2(v 1 w 1 +v 2 w 2 +v 3 w 3 )
=‖v‖ 2 +‖w‖ 2 −2(v·w) , so
−2‖v‖‖w‖cosθ=−2(v·w) , so since v0and w0then
cosθ= v·w , since‖v‖>0and‖w‖>0. QED
‖v‖‖w‖
Example 1.5. Find the angleθbetween the vectors v=(2,1,−1) and w=(3,−4,1).
Solution: Since v·w=(2)(3)+(1)(−4)+(−1)(1)=1,‖v‖= √ 6, and‖w‖= √ 26, then
cosθ= v·w
‖v‖‖w‖ = 1
√
6
√
26
=
1
2 √ ≈ 0.08 =⇒ θ=85.41◦
39
Two nonzero vectors are perpendicular if the angle between them is 90 ◦ . Since
cos90 ◦ = 0, we have the following important corollary to Theorem 1.6:
Corollary 1.7. Twononzerovectorsvandwareperpendicularifandonlyifv·w=0.
We will write v⊥wto indicate that v and w are perpendicular.