The Distance from a Point to a Plane

The shortest route

The distance from a point P to a plane is the length of the shortest route from P to the plane. That route runs along the normal: drop a perpendicular from P, and call the point where it meets the plane F, the foot of the perpendicular.

Any other point Q of the plane is farther away. The triangle PFQ has a right angle at F, because PF is perpendicular to every line in the plane, including FQ. So PQ is the hypotenuse, and a hypotenuse is longer than either other side.

P at the top, the foot F at the right angle and another point Q of the plane, with the plane running along the bottom edge. The perpendicular PF = 4 is a leg; the route PQ = 5 is the hypotenuse and is longer.

The part of the gap along the normal

Take any point A of the plane, with position vector a, and the gap from A up to P, which is p − a. The distance from P to the plane is the part of that gap that runs along the normal: its length in the direction of the unit normal n ÷ |n|.

That part is (p − a) · n ÷ |n|. Since A is on the plane r · n = d, a · n = d, so (p − a) · n = p · n − a · n = p · n − d. The distance is |p · n − d| ÷ |n|, with the modulus because a distance is never negative.

In Cartesian form this is a simple recipe. For the plane $n_{1}x+n_{2}y+n_{3}z=d$n₁x + n₂y + n₃z = d and the point P, substitute P into the left-hand side, subtract d, take the size, and divide by the length of the normal.

The slice through P(3, 3, 3), the point A(3, 0, 0) of the plane 2x + y + 2z = 6, and the normal, with the plane seen edge-on. The gap p − a = (0, 3, 3) has length $3\sqrt{2}$3√2, and its part along the unit normal is 3, the distance PF. Its part along the plane, AF, is also 3.

Read the white arrow a as the direction of the normal and the gold arrow b, of length 3, as the gap from a point of the plane to P. The green bar is the shadow of b on the line of a, $|b|\cos \theta$|b| cos θ: the part of the gap along the normal, which is P’s distance from the plane. Swing b to 90° and the gap lies along the plane, and the distance is 0.

A worked distance

Find the distance from P(3, 3, 3) to the plane 2x + y + 2z = 6. The normal is n = (2, 1, 2), with length $\sqrt{4+1+4}=3$√(4 + 1 + 4) = 3.

Substitute P into the left-hand side: 2 × 3 + 3 + 2 × 3 = 15. Subtract the constant: 15 − 6 = 9. Divide by the length of the normal: 9 ÷ 3 = 3. P is 3 units from the plane.

Check by projection. A(3, 0, 0) is on the plane, since 2 × 3 = 6. The gap is p − a = (0, 3, 3), and (0, 3, 3) · (2, 1, 2) ÷ 3 = (0 + 3 + 6) ÷ 3 = 3.

The foot of the perpendicular

The foot F is reached by walking from P along the normal, toward the plane, for the distance 3. The unit normal is (2, 1, 2) ÷ 3, so 3 units of it is exactly (2, 1, 2). P gave a positive reading, 9, so it lies on the side the normal points to, and the walk goes against n: F = (3, 3, 3) − (2, 1, 2) = (1, 2, 1).

Check: 2 × 1 + 2 + 2 × 1 = 6, so F is on the plane.

The foot can also be found as the meeting point of a line and a plane. The line through P along the normal is $r=\left(3,3,3\right)+\lambda \left(2,1,2\right)$r = (3, 3, 3) + λ(2, 1, 2). Substituting it into the plane gives $2\left(3+2\lambda \right)+\left(3+\lambda \right)+2\left(3+2\lambda \right)=6$2(3 + 2λ) + (3 + λ) + 2(3 + 2λ) = 6, which is $15+9\lambda =6$15 + 9λ = 6, so $\lambda =-1$λ = −1 and F = (3, 3, 3) − (2, 1, 2) = (1, 2, 1) again.

Which side of the plane?

Put the origin into the same plane: 2 × 0 + 0 + 2 × 0 − 6 = −6. The distance is |−6| ÷ 3 = 2. The minus sign only says which side of the plane the origin is on.

P gave +9 and the origin gave −6. Opposite signs mean opposite sides, so the segment from the origin to P crosses the plane. Two points that give the same sign are on the same side.

Two parallel planes

The distance between two parallel planes is the distance from any point of one to the other. P(3, 3, 3) lies on 2x + y + 2z = 15, since 6 + 3 + 6 = 15, so the planes 2x + y + 2z = 15 and 2x + y + 2z = 6 are |15 − 6| ÷ 3 = 3 apart.

When the normal’s length is not a whole number, the method is the same. The point (1, 1, 1) and the plane x + y + z = 1 give $|1+1+1-1|÷\sqrt{3}=\frac{2}{\sqrt{3}}$|1 + 1 + 1 − 1| ÷ √3 = 2/√3, about 1.15.

The usual mistakes

Leaving out the division. 9 is measured in whole normals, each 3 long; the distance is 9 ÷ 3 = 3.

Forgetting to subtract the constant. The substitution alone, 15, is not the gap; the plane’s 6 has to come off.

Measuring straight down. Dropping vertically only gives the distance when the plane is level; for a tilted plane the vertical route meets it at a slant and is longer.

Walking the wrong way, or too far, to the foot. From P, (3, 3, 3) + (2, 1, 2) = (5, 4, 5) moves away from the plane, and subtracting three whole copies, (3, 3, 3) − 3(2, 1, 2) = (−3, 0, −3), goes 9 units instead of 3.

A cable to the ceiling

In the application below, a sloping loft ceiling is a plane and a router is a point below it. The distance formula gives the length of the shortest cable, and the line from the router along the normal, substituted into the ceiling, gives where the cable meets it.