Calculus: Analytic Geometry and Calculus, with Vectors

2.3 Scalar products, direction cosines, and tines in E, 67

2.3 Scalar products, direction cosines, and lines in E3 Let u and v

be vectors in Ea having scalar components ul, u2, u3 and vi, os, v3 so that

(2.31)

IV

u = uli + uj + uak

= vii + rJ + oak.

It may be helpful to look at Figure 2.311, which shows vectors u and v

and the projections of their tips on the xy plane.

Figure 2.311

Since the scalar product is defined by the formula

(2.312) Jul Ivl cos 0,

it could be supposed that we should find cos 8 in order to find It

turns out, however, that there is a very simple and useful formula for

and we can use this formula to find cos 8 whenever we want it because we

know that

(2.313) lul=1/u;+ui+ua, IVl= o21 +v2+va.

We find that

(uli + utJ + o2j + oak)

= o2j + v,k) + v2j + oak) + ozj + oak).

With the aid of (2.231), we see that this reduces to the very important

formula

(2.32) ulvl + u2o2 + uaoa.

In order to help remember this formula, we can remember that the scalar

product of two rectors is the sum of the products of their scalar components.

Use of (2.32) and (2.312) gives, when Jul lvl ?d 0, the formula

(2.321) cos 0 = ulyi + u2v2 + U3V3

Jul IvI

which gives the angle 0 between two vectors in terms of the scalar com-
ponents of the vectors. This formula may be remembered. It is, in the