Calculus: Analytic Geometry and Calculus, with Vectors

72 Vectors and geometry in three dimensions

lies on the line PiP2 and hence that there is a number X for which P1P = AP1P2i
show how vectors are used to obtain the coordinate equation

x - x1 y - Y1 z - z1

x2-xl Y2 -Y1 z2 -zl

Refer to Figure 2.36 and the text only if assistance is needed.

9 Show that the equations

(1) x - x1 = X(x2 - xl), ``

Y - y1 = X(Y2 - Yl), z - z1 = n(z2 - zl),

from which the text of this section obtained the equations

(2)

x-x1 - Y -Y1 -z-z1 1

x2-x1 Y2-Y1 z2-z1

can be put in the form

(3) x = (x2 - xl)X + xi, Y -- (y2 - yl)X + y1,

Then show that the equations

z= (z2- z1)X + Z1-

(4) x = a1X + b1, y = a2X + b2, z = aaX + bs

can be put in the form

(5) al

a2 a3

Remark: The equations in (1), (3), and (4) are called parametric equations of

lines; different values of the parameter X yield different points on the lines. It is

fashionable to use the letter X for a parameter because people who work with

Lagrange (French mathematician) parameters (or multipliers) get the habit;

the letter i is used when time is involved and sometimes when time is not involved.

10 Look at the equations

x-x1 y-Y1 z- 2:1

x2-x1-Y2Yi-z2z1

of the line L which passes through the points P1(x1,yl,zl) and P2(x2,y2,z2) and tell

why these equations are satisfied when x = x1, y = y1, z = z1 and when x = x2,

Y = y2, z = z2. With this is mind tell how we can quickly find the coordinates

of two points on the line having equations

x-2 y+3 z+1

1-2=2+3-3+1

(^11) Find equations of the line L which passes through the points P1(2,-3,-1)
and P2(1,2,3) and then, after putting z = 0, find the coordinates of the point

(x,y,0) in which L intersects the xy plane. Ans.: x = y = -1 z = 0.

(^12) This problem involves a little introductory information that everyone
should have. The right-handed coordinate system shown in Figure 2.391 is the
one we ordinarily use. When we are wholly or primarily interested in vectors