Multivariate Calculus - Bruce E. Shapiro

14 LECTURE 2. VECTORS IN 3D
Theorem 2.5 u · v = ||u‖‖v‖ cos ϑ, where ϑ is the angle between the two vectors u
and v.
Proof. We can use the law of cosines to determine the length ‖u − v‖ (see figure
2.5).
Figure 2.5: The triangle with sides ‖u‖, ‖v‖, and ‖u − v‖.
Hence
‖u − v‖ 2 = ‖u‖ 2 + ‖v‖ 2 − 2‖u‖‖v‖ cos θ
2‖u‖‖v‖ cos θ = ‖u‖ 2 + ‖v‖ 2 − ‖u − v‖ 2
= u · u + v · v − (u − v) · (u − v)
= u · u + v · v − [u · (u − v) − v · (u − v)]
= u · u + v · v − [u · u − u · v − v · u + v · v]
= 2 u · v
Therefore the dot product can be written as
u · v = ‖u‖‖v‖ cos θ
Corollary 2.1 . Two vectors are perpendicular if and only if their dot product is
zero.
Example 2.3 Find the angle between the two vectors in ⃗u = (1, −3, 7) and ⃗v =
(16, 4, 1)
Solution. We use the fact that u · v = ‖u‖‖v‖ cos θ. From the previous examples,
we have
‖u | = √ 59
‖v‖ = √ 273
u · v = (1)(16) + (−3)(4) + (7)(1) = 16 − 12 + 7 = 11.
Revised December 6, 2006. Math 250, Fall 2006