Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1
2.4 Planes and lines in E, 85

and a given point Po(xo,yo,zo). Hint: Consider separately the case in which Po
is on L and the case in which Po is not on L.
24 Let 7r1 and 72 be the planes having the equations


J2x+3y+4z-5 = 0
(1) l x-2y+3z-4=0

Verify that 1r1 and ire do not have parallel normals. This implies that al and 7r2
must intersect in a line L. Observe that a point P(x,y,z) lies on L if and only if it
lies on both al and a2 and hence if and only if its coordinates satisfy both of the
equations (1). We should be able to find point-direction equations of L by
finding the coordinates of two points P, and P2 on L and using them. Do this by
finding x and y such that the point (x,y,O) lies on L and then finding x and y such
that the point (x,y,l) lies on L. We now develop a simpler and more interesting
method for finding equations of L. Show that if P(x,y,z) lies on L, then
(2) 7x + 17z - 22 = 0.
Observe that (2), the result of eliminating y from the equations (1), is obtained
by multiplying the equations (1) by 2 and 3, respectively, and adding the results.
Observe that (2) is the equation of the plane which passes through L and is
perpendicular to the xz plane. Show also that if P(x,y,z) lies on L, then

(3) 7y-2z+3=0.
Discuss this matter. By solving the equations (2) and (3) for z, show that

(4) 7x - 22_7y + 3
-17 - 2

and that these are equations of L. Observe that dividing these equations by 7
puts them in the point-direction form
22
(5)

x- =Y+3-z-0
-17 2 7

25 Find the equation of the plane in which contains the point (1,3,1) and the
line L having the equations
(1) x=t, y=t, z=t+2.

First solution: The equation of in has the form

(2) 4(x - 1) + B(y - 3) + C(z - 1) = 0.


Since in contains the given line, we must have
(3) .4(t - 1) + B(t - 3) + C(t + 1) = 0

for each t. Putting t = 3 and then t = 1 shows that

(4) 2.4+4C=0, -2B+2C=0.

Solving these equations for .4 and B and putting the results in (2) gives

(5) -2C(x - 1) + C(y - 3) + C(z - 1) = 0.

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