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2 Intersecting 3-D planes

Attachment 22508

I have been at this same problem for a few hours now. The scan of the paper is above. I'm trying to find the parametric and symmetric equations for the line of intersection for the planes:

3x - 3y - 7z = -4 and x - y + 2z = 3.

I thought I had it solved but if i plug the parametric equations into the original equations neither comes out right. It would if one of the signs where different for the 13t in the parametric equations. But I don't know where I dropped it.

Please help find my mistake or what I'm doing wrong.

Thank you.

Re: 2 Intersecting 3-D planes

nvm I messed up on the sign in the cross products.

Re: 2 Intersecting 3-D planes

Quote:

Originally Posted by

**DerekZ10** Attachment 22508
equations for the line of intersection for the planes:

3x - 3y - 7z = -4 and x - y + 2z = 3.

I thought I had it solved but if i plug the parametric equations into the

Can you show that the point $\displaystyle (1,0,1)$ is on both planes?

The direction vector is: $\displaystyle <3,-3,-7>\times<1,-1,2>.$

Re: 2 Intersecting 3-D planes

I would think that a rather obvious thing to do would be to subtract 3 times the second equation from the first:

(3x- 3y- 7z)- 3(x- y+ 2z)= -4- 3(3)

-13z= -13 so that z= 1, a constant.

With z= 1, the two equations become 3x- 3y- 7= -4 and x- y+ 2= 3 which both reduce to x- y= 1.

x= y+ 1. Using y as parameter, x= t+1, y= t, z= 1 are parametric equations for the line of intersection.

Re: 2 Intersecting 3-D planes

Is there possibly 2 answers? I went with x = 1 - 13t, y = -13t, and z = 1. Substitution of those worked on both equations. My original mistake was that I dropped neg sign on y = 13t.