# Thread: Cartesian eqn of a plane

1. ## Cartesian eqn of a plane

A plane passes through the P, with position vector i + 2j - k, and is perpendicular to the line L with eqn r = 3i -2k + w(-i + 2j + 3k)

Show that the Cartesian eqn of the plane is x - 5y -3z = -6

I'm not sure how to work out the normal of the plane?

2. Originally Posted by Erghhh
A plane passes through the P, with position vector i + 2j - k, and is perpendicular to the line L with eqn r = 3i -2k + w(-i + 2j + 3k)
Show that the Cartesian eqn of the plane is x - 5y -3z = -6
That given answer is answer is not perpendicular to the line L.
The answer I get is $x-2y-3z=0$.

3. Yeah, I was careful. It was from a past paper and I thought there was a typo. And on the mark scheme it says that the normal was 1i + 5j + 3k. I was skeptical about there being two typos on two separate documents. But I do agree with your answer.