Calculus: Analytic Geometry and Calculus, with Vectors

86 Vectors and geometry in three dimensions

Dividing by C and multiplying by -1 gives

(6) 2(x-1)-(y-3)-(z-1)=0

or

(7) 2x-y-z+2=0.

Second solution: Eliminating t from the first two and then from the first and last

of the equations (1) shows that L is the intersection of the planes having the

equations

(8) x-y=0, x-z+2=0.

Each plane containing L has an equation of the form

(9) X(x-y)+μ(x-z+2) =0,

where X and μ are constants. The plane having the equation (9) contains the

point (1,3,1) if and only if -2X + 2μ = 0. Putting μ = X gives the required

equation

(10) 2x-y-z+2=0.

26 There are nontrivial applications of the idea that a point which lies in

each of two nonparallel planes must lie on their line of intersection. Show that

the equation

x2 y2 z2

a2+ b2 62 = 1

will be satisfied if, for some constant X, the two equations

a+c=x(1+b/' x\a( e)-1 b

both hold. Try to determine a geometrical interpretation of this result. Try

to obtain another very similar result.

27 Let L and L' be the lines of intersection of the planes having the equations

a+c-all+b/

L

Ca cJ =1 - b

L'

z

μ )=l+b

Work out the point-direction forms

2Xa (1 - X2)b

L:

f+ X2 1+X2 - z-0

-(1 - X2)a 2Xb (1 + X2)c

x _ 2μa (1 - μ2)b

1+μ2 y+ 1+μ2 _ z-0

-(1 - μ2)a -2μb (1 + μ2)c

of equations of L and L'.