I realize this may be VERY low level for this forum. I'm practicing for an exam and I just want to verify an answer because I do not have the solutions for this practice test.
The question is:
Find an equation of the plane containing $(1, 5, -1)$ that is perpendicular to the line of intersection of the two planes: $– x + y – 5z = 4$ and $2x – y – 2z = 0$
Here is my solution:
I found the intersect of the two planes and found the equations:
$ x = 7t + 4 $
$ y = 12t + 8 $
$ z = t $
I used z = t as a free variable.
Then I used $n = (7, 12, 1)$ as normal vector for the desired plane and $P = (1, 5, -1)$ for the point.
Ended up having: $7(x - 1) + 12(y - 5) + (z + 1) = 0$ and $7x + 12y + z = 66$
can anyone confirm this answer for me please? thank you