Problem: Show that the line $x – 2 = \frac{y+3}{2} = \frac{z-1}{4}$ is parallel to the plane $2y – z = 1$. What is the distance between the line and the plane?
Attempt at a solution: I'm not sure how to show they are parallel; I'm confused by the way the line is expressed. But if somehow we could show this, then we could specify a point on that line by setting $y=z=0$ and then finding the distance from that point to the plane by the formula \begin{align*} \frac{|A x_0 + B y_0 + C z_0 – D|}{\sqrt{A^2 + B^2 + C^2}}\end{align*}
Any help in showing they are parallel would be appreciated.
Best Answer
A normal vector to the plane is $(0,2,-1)$.
A direction vector for the line is $(1, 2, 4)$.
These vectors are orthogonal, so the is line parallel to the plane.
To get the distance, select a point $P$ on the line and a point $Q$ on the plane: $P = (2,-3,1)$ and $Q = (0,0,-1)$ are natural choices. Just compute the scalar projection of $PQ$ onto the normal vector $n = (0,2,-1)$.