# Math: Intersections of 3 Planes

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.

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1. 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

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2. let l be th line x+1/2=y+3/3=-z and let P be the plane 3x-2y+4z=-1. find the piont of intersection po of l and p

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