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## Homework Statement

Find the parametric equation for a line of intersection of these two planes

x+2y+3z=0

4x+5y+6z=5

## Homework Equations

Normal to plane 1= <1,2,3>

Normal to plane 2= <4,5,6>

## The Attempt at a Solution

I know the way to do this problem is to take cross product of two normals etc etc,

but i want to know if the way i did this is correct also.

I already turned the work in so theres nothing i can do to change it but the curiosity is killing me.

First, i set the two planes equal to each other

4x+5y+6z-5=0

x+2y+3z=0

=> x+2y+3z=4x+5y+6z-5 (please correct me my thinking process is wrong,im winging it)

=> 3x+3y+3z=5

=> x+y+z=5/3 (now im thinking this is the equation of the intersection of the two planes but this isnt the equation of a line, it looks like a plane, or is it?)

so i took a point on this set, (5/3,0,0) and two other points (0,5/3,0) and (0,0,5/3)

i did <0,0,5/3> - <0,5/3,0> = <0,-5/3,5/3> as a directional vector.

so L(t)= (5/3,0,0) + <0,-5/3,5/3>t

x=5/3

y=-5/3t

z=5/3t

IS any of this wrong?!