What is the distance between the line and plane if it is parallel?

Question

So far, I've gotten that the line is parallel to the plane $x = 2 + t$, $y = -3 + 2t$, $z = 1 + 4t$ With the vector of that being $U$ is $(1,2,4)$ and the plane $2y-z = 1$ with the vector $V$ being $(0,2,-1)$.

I'm kind of stuck because I'm not entirely sure where to go from here, any tips or advice would be greatly appreciated!

Answer 1

This question can be reduced by calculate the distance from a point to a line (or plan): You have to take a point A ((2, -3, 1) for example) from line and a point H ((0, 0, -1) for example) from plan.

Here $\overrightarrow{AH} = (-2, 3, -2)$

As you know normal of plan : $\overrightarrow n = (0, 2, -1)$, distance is : $$d(H, (d)) = d(A, (P)) = \frac{\|\overrightarrow{AH} \dot{} \overrightarrow n\|}{\|\overrightarrow n\|}$$

Answer 2

Hint: can you calculate the distance from a point to a plane (use the normal to the plane). Pick any two points on the line. If the distance to the plane is the same, the line is parallel to the plane, and the distance is what you calculate.