Foot-of-Perpendicular Method

The foot of the perpendicular from a point P to a line or plane is the point F of that object closest to P; it is found by writing F in terms of a parameter and requiring that PF be perpendicular to the direction of the line (PF · d = 0) or parallel to the normal of the plane (F = P − (f(P)/|n|²) n). In brief: the foot of the perpendicular from P(2, 3) to the line 3x + 4y − 10 = 0 is F = (2, 3) − (8/25)(3, 4) = (26/25, 43/25), and |PF| = 8/5 is the distance.

The foot-of-perpendicular method is a basic tool for distances, projections and reflections. This guide treats the foot on a line in the plane, on a line in space, on a plane, and the common perpendicular of two skew lines, with worked examples and a table of formulas.

What is the idea of the method?

Among all points X of a line or plane, the closest one to P is the foot F, and PF is perpendicular to the object. This is the geometric content of the Pythagorean theorem: for any other point X, |PX|² = |PF|² + |FX|². The method has two steps: parametrize the object and use the perpendicularity condition as an equation for the parameter.

How do you find the foot on a line?

For the line X(t) = A + t d, the foot corresponds to t = ((P − A) · d)/(d · d). Then F = X(t) and the distance is |P − F|.

Example 1: a line in the plane in general form

P(2, 3), line 3x + 4y − 10 = 0 with the normal n = (3, 4). The foot is F = P − (f(P)/|n|²) n with f(P) = 6 + 12 − 10 = 8: F = (2, 3) − (8/25)(3, 4) = (26/25, 43/25). Check: 3 · 26/25 + 4 · 43/25 = 250/25 = 10 ✓. The distance is (8/25) · 5 = 8/5.

Example 2: a line in the plane in parametric form

P(0, 5), line (0, 0) + t(1, 1). t = (P · d)/(d · d) = 5/2, so F = (2.5, 2.5) and the distance is √(6.25 + 6.25) ≈ 3.54.

Example 3: a line in space

P = (0, 0, 0), line A(1, 2, 3) + t(3, −2, 2). (P − A) · d = (−1, −2, −3) · (3, −2, 2) = −3 + 4 − 6 = −5, d · d = 17, so t = −5/17. F = (1 − 15/17, 2 + 10/17, 3 − 10/17) = (2/17, 44/17, 41/17). Check: F · d = (6 − 88 + 82)/17 = 0 ✓. The distance is √(4 + 1936 + 1681)/17 = √3621/17 ≈ 3.54, the same as the cross-product formula √(213/17).

How do you find the foot on a plane?

For the plane n · X + D = 0 and the point P, let f(P) = n · P + D. The foot is F = P − (f(P)/|n|²) n, because the segment PF runs along the normal. The distance is |f(P)|/|n|. Alternatively, intersect the line P + s n with the plane.

Example 4: a foot on a plane

P(1, 2, 3), plane 2x − 2y + z − 9 = 0, n = (2, −2, 1), |n|² = 9, f(P) = 2 − 4 + 3 − 9 = −8. F = (1, 2, 3) + (8/9)(2, −2, 1) = (25/9, 2/9, 35/9). Check: 2 · 25/9 − 2 · 2/9 + 35/9 = 81/9 = 9 ✓. The distance is 8/3.

Example 5: by intersecting the normal line

The same problem: the line (1, 2, 3) + s(2, −2, 1) enters the plane when 2(1 + 2s) − 2(2 − 2s) + (3 + s) − 9 = 0, so 9s − 8 = 0 and s = 8/9, the same F.

Example 6: the projection of a line onto a plane

The projection of a line is found by projecting two of its points. The line through (0, 0, 0) and (1, 1, 1) on the plane z = 0 projects to the points (0, 0, 0) and (1, 1, 0), that is the line y = x, z = 0. In general, project the point and the direction.

How do you find the common perpendicular of skew lines?

For L₁ = A₁ + t d₁ and L₂ = A₂ + s d₂, the common perpendicular joins the points P₁(t) and P₂(s) with (P₁ − P₂) · d₁ = 0 and (P₁ − P₂) · d₂ = 0. These are two linear equations in t and s.

Example 7: the shortest segment between skew lines

L₁: (1, 2, 3) + t(2, 1, −1), L₂: (4, 1, 3) + s(1, 1, 1). P₁ − P₂ = (−3 + 2t − s, 1 + t − s, −t − s). The first condition (dot with (2, 1, −1)): −5 + 6t − 2s = 0. The second (dot with (1, 1, 1)): −2 + 2t − 3s = 0. From the second, t = (2 + 3s)/2; then 3(2 + 3s) − 2s = 5, so s = −1/7 and t = 11/14. P₁ = (18/7, 39/14, 31/14), P₂ = (27/7, 6/7, 20/7). P₁ − P₂ = (9/14)(−2, 3, −1), of the length (9/14)√14 = 9/√14 ≈ 2.41. The direction (−2, 3, −1) is parallel to d₁ × d₂ = (2, −3, 1), as expected.

Table of formulas

Where does the idea come from?

The fact that the shortest segment from a point to a line is the perpendicular is found in Euclid’s Elements (the theory of perpendiculars and the Pythagorean relation). The orthogonal projection became central in the method of least squares, introduced by Legendre (1805) and Gauss, which minimizes distances in exactly this sense.

Which mistakes are common?

Forgetting to divide by d · d

t = (AP · d)/(d · d). If d is a unit vector the denominator is 1; otherwise, omitting it gives the wrong foot.

Using AP with the wrong sign

AP = P − A, not A − P. A sign error moves the foot to the other side of A.

Mistaking the normal for the direction

For a plane, the foot lies along the normal; for a line in space, the foot lies in the plane perpendicular to d. Do not exchange the roles.

Skipping the check

Always test that PF · d = 0 (or PF ∥ n) and that F lies on the object.