# Parametric equation for a line which lies on a plane

Struggling to begin answering the following question:

Let $L$ be the line given by $x = 3-t, y= 2+t, z = -4+2t$. $L$ intersects the plane $3x-2y+z=1$ at the point $P = (3,2,-4)$. Find parametric equations for the line through $P$ which lies on plane and is perpendicular to $L$.

So far, I know that I need to find some line represented by a vector $n$ which is orthogonal to $L$. So, with the vector of $L$ represented by $v$, I have:

$$n\cdot v = 0 \Rightarrow [a, b, c] \cdot [-1, 1, 2] = 0 \Rightarrow -a + b + 2c = 0$$

I am not sure how to proceed from here, or if I am even on the right track.

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I think you mean to say that the line lies on the plane and is perpendicular to $L$. – Michael Kasa May 4 '12 at 3:55
Sorry, typographical error. Will fix. – Dylan May 4 '12 at 3:58

You know that the line you want is perpendicular to the line L, which has direction vector $\langle -1,1,2\rangle$, and that the line you want lies in the given plane, which has normal vector $\langle 3, -2, 1\rangle$. So the line you want is orthogonal to both $\langle -1,1,2\rangle$ and $\langle 3, -2, 1\rangle$ and you can use the cross product of these two vectors as the direction vector of the line you want.
I am not sure if this is right, but once I have found the vector resulting from the cross product, $[5, 7, 1]$, I need to use the above point $P(3, 2, -4) to put into parametric form, correct? – Dylan May 4 '12 at 4:28 @Dylan: I get$\langle 5, 7, -1\rangle$(positive 7, not negative 7), but I didn't double-check my work. Yes, with a known point$P$and a direction vector$v$, the line is$X=P+vt\$. – Isaac May 4 '12 at 4:30