# Showing a line is parallel to a plane and finding the distance between them

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.

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 $n = (0,2,-1)$.

• How did you derive the direction vector? Mar 3, 2015 at 20:14
• The standard form for a line is $\dfrac{x-x_0}{a} = \dfrac{y-y_0}{b} = \dfrac{z - z_0}{c}.$ The direction vector is just $(a,b,c)$. Mar 3, 2015 at 20:15
• Thanks for the help. Did you conclude they are orthogonal because one is not a multiple of the other? Mar 3, 2015 at 20:20
• Just compute the dot product: it is zero. Mar 3, 2015 at 20:21
• I see. And for the distance, can I pick an arbitrary point? Like if I set $y=z=0$. Never mind: that wouldn't work. Equation would have a contradiction. Mar 3, 2015 at 20:28

Two possible methods for arriving at a solution.

(1) For a line and a plane in 3 dimensions there are only 3 possibilities: (a) the line lies in the plane; (b) the line and plane intersect at a single point; or (c) the line and plane are parallel. So one thing you can do is eliminate the other possibilities.

(2) They will be parallel if the normal to the plane is also perpendicular to the tangent of the line.