Math: Intersections of 3 Planes
posted by Anonymous on .
For what value of "k" will the following set of planes intersect in a line?
x  2y  z = 0
x + 9y  5z = 0
kx  y + z = 0
Work:
Elimination to find "z":
x  2y  z = 0
x + 9y  5z = 0

11y + 4z = 0
z = 11y/4
Sub "z" into "x  2y  z = 0" to find "x":
x  2y (11y/4) = 0
x  19y/4 = 0
x = 19y/4
Sub "x" and "z" into "x + 9y  5z = 0" to find "y":
(19y/4) + 9y  5(11y/4) = 0
(19y + 36y  55y) / 4 = 0
0y/4 = 0
So frustrated...I don't know what to do for this question. Please help me (step by step solution).
Thanks.

Find the intersection of the first two lines:
subtract the first from the second
11y  4z = 0
11y = 4z
y/z = 4/11
let y=4 and z=11 (it satisfies the above ratio)
back in the first
x  8  11 = 0
x = 19
since clearly (0,0,0) lies on all three planes, the line joining (0,0,0) and (19,4,11) is the intersection line of the first two planes.
But all 3 planes are supposed to intersect in the same line, so obviously (19,4,11) must also lie on the third plane.
so sub it in....
k(19)  8  11 = 0
k = 19/19 =1
so k=1 
let l be th line x+1/2=y+3/3=z and let P be the plane 3x2y+4z=1. find the piont of intersection po of l and p