Recipe

Find the distance from a point to a line or a plane

Use when: you have a point not on a given line or plane and need the shortest distance from it

Goal: a single nonnegative distance value, from whichever of the two formulas matches what the point is being measured against

Taught in Distances between points, lines, and planes

Quick recipe

Decide whether the target is a line or a plane
For a plane, read off the normal and plug into the formula
For a line, find a vector from the line's point to $P_1$
Cross that vector with the direction, then divide by $|\mathbf v|$

1
Read the problem to see which kind of object the point's distance is being measured against.
- distance to a plane — continue to Step 2.
- distance to a line — skip to Step 4.
Why: the two targets use different formulas built from different products, a dot product for a plane and a cross product for a line, so branching here keeps them from getting mixed up.
2
Read $\mathbf n=\langle a,b,c\rangle$ straight off the plane's equation $ax+by+cz=d$ , then compute $D=\dfrac{|ax_1+by_1+cz_1-d|}{|\mathbf n|}$ for the point $P_1=(x_1,y_1,z_1)$ .

Example

Plane $2x-y+2z=3$ , point $P_1=(4,-3,5)$ : $|\mathbf n|=\sqrt{4+1+4}=3$ , and $2(4)-(-3)+2(5)-3=8+3+10-3=18$ , so $D=18/3=6$ .

Why: this is the scalar projection of a vector from any plane point to $P_1$ onto the unit normal, with the absolute value stripping away which side of the plane $P_1$ happens to sit on.
3
Compute $\overrightarrow{P_0P_1}$ by subtracting the line's known point from $P_1$ .

Example

Line through $P_0=(1,1,1)$ with direction $\mathbf v=\langle-2,-2,-1\rangle$ , point $P_1=(-1,2,3)$ : $\overrightarrow{P_0P_1}=\langle-2,1,2\rangle$ .

Why: the cross-product formula in the next step needs this vector, not $P_1$ 's coordinates alone.
4
Compute $D=\dfrac{|\overrightarrow{P_0P_1}\times\mathbf v|}{|\mathbf v|}$ .

Example

$\overrightarrow{P_0P_1}\times\mathbf v=\langle3,-6,6\rangle$ , with magnitude $9$ . $|\mathbf v|=\sqrt{4+4+1}=3$ , so $D=9/3=3$ .

Why: the cross product's magnitude is the area of the parallelogram $\overrightarrow{P_0P_1}$ and $\mathbf v$ span, and dividing that area by the base $|\mathbf v|$ leaves the parallelogram's height, exactly the point's perpendicular distance to the line.

Common mistakes

Using the plane's formula for a line, or the line's formula for a plane.
Forgetting the absolute value in the point-to-plane formula and reporting a signed distance.
Building $\overrightarrow{P_0P_1}$ backward, from $P_1$ toward $P_0$ , which can flip a sign elsewhere without changing the final magnitude.
Dividing by $|\mathbf n|^2$ or $|\mathbf v|^2$ instead of the plain magnitude in the final step.