# Thread: Parametric equation of lin eof intersection of two planes?

1. ## Parametric equation of lin eof intersection of two planes?

Hi , I am doing this linear algebra question:
Find the parametric equations of the line of intersection of the planes x+2z=0 and 2x-3y+4
I don't know whether I am approaching this the correct way or if my answer is correct. What I did was found the cross product of the normals of the vectors which gave me (3,2,-3); then I subbed in zero for x to get 2z=0, -3y=4-> z=0, y=-4/3 and got the equation of the line as being L= (0,0,-4/3)+t(3,2,-3)
Any feedback would be greatly appreciated.

2. Originally Posted by macduff
Hi , I am doing this linear algebra question:
Find the parametric equations of the line of intersection of the planes x+2z=0 and 2x-3y+4
I don't know whether I am approaching this the correct way or if my answer is correct. What I did was found the cross product of the normals of the vectors which gave me (3,2,-3); then I subbed in zero for x to get 2z=0, -3y=4-> z=0, y=-4/3 and got the equation of the line as being L= (0,0,-4/3)+t(3,2,-3)
Any feedback would be greatly appreciated.
I'd just solve

x + 2z = 0 .... (1)

2x - 3y = 4 .... (2)

simultaneously:

Let x = t. Then y = (2t - 4)/3 and z = -t/2. This is the parametric equation of a line.

Your solution is not consistent with mine eg. My line does not pass through (0, 0, -4/3) but yours does (in fact, that point does not even lie in the second plane ....)

3. ## Thanks

Thankyou so much your way definitely makes more sense and is easier :P
Originally Posted by mr fantastic
I'd just solve

x + 2z = 0 .... (1)

2x - 3y = 4 .... (2)

simultaneously:

Let x = t. Then y = (2t - 4)/3 and z = -t/2. This is the parametric equation of a line.

Your solution is not consistent with mine eg. My line does not pass through (0, 0, -4/3) but yours does (in fact, that point does not even lie in the second plane ....)