Calculus: Analytic Geometry and Calculus, with Vectors

2.3 Scalar products, direction cosines, and lines in E, 69

which provides an interesting way of making the more prosaic statement

that the sum of the squares of the numerical components of a unit vector

is 1. It is possible to put the formula (2.321) in the form

(2.344) cos 0 = cos a, cos a2 + cos$1 cos 02 + cos 71 cos 72,

where 0 is the angle between two vectors making angles a,, $i, 'y, and

a2, 02, 'y2 with the unit vectors 1, j, k. The formulas (2.343) and (2.344)

are impressive formulas, but persons who know about vectors may hold

the view that the formulas are antiquated and may concentrate their

attention upon the formulas (2.33).

The following numerical example shows an application of the ideas of

this paragraph. The vector in the left member of the formula

21-3j+4k=1/29 2 1- (^3) j+ 4 k
1-1-1/29 )
is the vector running from the origin 0 to the point P(2,-3,4). The
length of the vector is the square root of the sum of the squares of its
numerical components and hence is. The vector in parentheses in
the right member is a unit vector in the direction of OP. The fact that
its scalar components are cosines of direction angles is very often of no
importance.
It is easy to adapt these ideas to obtain information about the vector
P1P2 running from P,(x,,y,,z1) to Ps(x2,y2,z2) as in Figure 2.36. Starting
with the formulas
(2.35) OP, = x1i + y,j + zlk, OP2 = x2i + y2j + z2k,
it is easy to see that the rules for addition and subtraction of vectors give
the formulas
OP2 + OP1 = (X2 + x1)i + (Y2 + Yi)j + (Z2 + z,)k
OP2 - OP1 = (x2 - xi)i + (Y2 - yi)j + (z2 - zi)k.
But OP2 - OP1 = P1P2, and hence
(2.351) P,P2 = (X2 - x,)i + (y2 - Y1)j + (z2 - zl)k.
Therefore,
(2.352) jP1PI = 1/(x2 - x1)2 + (Y2 - yl)2 + (z2 - z1)2.
If to simplify writing we let d denote this distance between P1 and P2, then
we can put (2.351) in /the form
\
(2.353) P,P2 = d ` x2dX I + Y2dY j + Z2 z1 k).