Multivariate Calculus - Bruce E. Shapiro

LECTURE 2. VECTORS IN 3D 15
Hence
u · v = ‖u‖‖v‖ cos θ
⇒ 11 = √ 59 √ 273 cos θ = √ 16107 cos θ
⇒ cos θ = √ 11 ≈ 11
16107 127 ≈ 0.087
⇒ θ ≈ arccos 0.087 ≈ 29 deg
Example 2.4 The basis vectors are all mutually orthogonal to one another:
i · i = j · j = k · k = 1
i · j = j · k = k · i = 0
Unit Vectors, Direction Angles and Direction Cosines
Definition 2.8 A unit vector is any vector with a magnitude of 1. If v is a unit
vector, it is (optionally) denoted by ˆv.
Theorem 2.6 Let v ≠ 0 be any vector. Then a unit vector ˆv parallel to v is
v
‖v‖
ˆv =
v
‖v‖
Proof. To verify that is a unit vector,
∥( v
∥∥∥ ∥‖v‖
∥ = vx
‖v‖ , v y
‖v‖ , v z ∥∥∥
‖v‖)∥
√ ( ) 2 (
vx vy
=
+
‖v‖ ‖v‖
1
√
= vx ‖v‖
2 + vy 2 + vz
2
To see that u =
v
‖v‖
= ‖v‖
‖v‖ = 1
and v are parallel,
u · v = v · v
‖v‖ = ‖v‖2
‖v‖ = ‖v‖
but, letting θ be the angle between u and v
u · v = ‖u‖‖v‖ cos θ = ‖v‖ cos θ
) 2 ( ) 2 vz
+
‖v‖
Equating the two expressions for u · v gives cos θ = 1 or θ = 1. Hence the vectors
are parallel. Since u has magnitude one and is parallel to v we conclude that
ˆv = u = v/‖v‖.
Math 250, Fall 2006 Revised December 6, 2006.