Question

Determine the intersection, if any, of the planes with equations x + y - z + 12 = 0 and 2x + 4y - 3z + 8 = 0.

I have no clue where to even begin. If someone could help and give me the steps to complete it, that would be greatly appreciated. Do I have to first simplify the equations? of find the normal vector for them? I am really stuck.

Answer 1

First of all you must understand what the intersection is
When you intersect two planes you get a straight line, unless the two planes are parallel
In this case they are not, since 1:1:-1 ≠ 2:4:-3

multiply the first by 3
3x+3x-3z = -36
subtract
2x+ 4y - 3z = -8
---------------
x - y = -28
x = y -28

pick any y, let y = 20
x = -8
in #1
-8 + 20-z = -12
z = 24 , so we have a point (-8 , 20 , 24)

another point: let y = 0
x = -28
-28 + 0 -z = -12
z = -16 , another point is (-28 , 0 , -16)

so the direction of our line is [20 , 20 , 40] or as a simplified vector [1,1,2]
using our first point of (-8,20,24), our line is

(x+8) = (y-2) = (z-24)/2

or as parametrics:
x = -28 + t
y = 20 + t
z = 24 + 2t

Answer 2

why should it multiply by 3