The line $L$ passes through the point $p = (1,-1,1)$ and has direction vector $d = [ 2,3, -1]$. Determine for the plane $P$, with equation $2x+3y-z = 1$ whether $L$ is parallel, perpendicular or neither.
So far I know I've been given a line in vector form, so my guess is to compare the normal line from the plane with the line. But I kind of stuck in that part, I don't know how to proceed with this problem, a hint or something is really appreciated.
Best Answer
For $L$ to be parallel to the plane $P$, all you have to do is checking the direction vector $\vec d$ of $L$ is orthogonal to a vector $\vec n$ normal to $P$, i.e. $\vec d \cdot \vec n=0$. You get such a vector from the equation of the plane
For $L$ to be perpendicular to $P$, check $\vec d$ is colinear to $n$, which you can do with the cross-product: $\vec d \times\vec n =\vec 0$.