Distance from a Point to a Line

Introduction

A point and a line in three-dimensional space are usually not enough information to determine a distance by simply subtracting coordinates: the shortest connecting segment may not point in the same direction as the line. The line’s direction matters because it tells us which part of the displacement from the line to the point is parallel to the line and which part is perpendicular.

The cross product gives a direct way to isolate the perpendicular contribution. Its magnitude measures the area of a parallelogram, so dividing that area by the length of the line’s direction vector produces the shortest distance.

Representing a Line and a Point

Write a line \(L\) in point-direction form as \(\mathbf{r}=\mathbf{a}+t\mathbf{v}\), where \(\mathbf{a}\) is the position vector of a point \(A\) on the line, \(\mathbf{v}\) is a nonzero direction vector, and \(t\) ranges over the real numbers. Let \(Q\) be the point whose distance from the line is required, with position vector \(\mathbf{q}\).

Here \(\mathbf{q}-\mathbf{a}\) is the displacement from the chosen point \(A\) on the line to \(Q\). The denominator is nonzero because the direction vector of a line must be nonzero. The result is a nonnegative length.

Why the Cross-Product Formula Works

Let \(\mathbf{d}=\mathbf{q}-\mathbf{a}\). Split this displacement into a component parallel to the line and a component perpendicular to it: \(\mathbf{d}=\mathbf{d}_{\parallel}+\mathbf{d}_{\perp}\). The parallel component is the projection of \(\mathbf{d}\) onto the nonzero vector \(\mathbf{v}\): \[ \mathbf{d}_{\parallel}=\frac{\mathbf{d}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\mathbf{v}, \qquad \mathbf{d}_{\perp}=\mathbf{d}-\mathbf{d}_{\parallel}. \] By construction, \(\mathbf{d}_{\perp}\cdot\mathbf{v}=0\).

The point on the line with parameter \(t_*=(\mathbf{d}\cdot\mathbf{v})/(\mathbf{v}\cdot\mathbf{v})\) has position vector \(\mathbf{a}+t_*\mathbf{v}\). The displacement from this point on the line to \(Q\) is exactly \(\mathbf{d}_{\perp}\), so its length is \(\lVert\mathbf{d}_{\perp}\rVert\).

It remains to relate this perpendicular length to a cross product. Since \(\mathbf{d}_{\parallel}\) is a scalar multiple of \(\mathbf{v}\), its cross product with \(\mathbf{v}\) is zero. Consequently, \(\mathbf{d}\times\mathbf{v}=\mathbf{d}_{\perp}\times\mathbf{v}\). The magnitude of the resulting cross product is the length of the perpendicular component multiplied by the direction-vector length: \[ \lVert\mathbf{d}\times\mathbf{v}\rVert = \lVert\mathbf{d}_{\perp}\rVert\lVert\mathbf{v}\rVert. \] This follows from the cross-product magnitude identity \(\lVert\mathbf{x}\times\mathbf{y}\rVert^2 =\lVert\mathbf{x}\rVert^2\lVert\mathbf{y}\rVert^2-(\mathbf{x}\cdot\mathbf{y})^2\), applied to perpendicular vectors. Dividing by \(\lVert\mathbf{v}\rVert\) proves the formula.

The cross product does not point in the direction of the perpendicular displacement. It points perpendicular to both vectors being crossed. It is its magnitude that gives the perpendicular distance after division by \(\lVert\mathbf{v}\rVert\).

Applying the Formula

When the Distance Is Not an Integer

Using the Formula with a Different Direction

A direction vector can be multiplied by any nonzero scalar without changing the line. The formula remains unchanged: multiplying \(\mathbf{v}\) by a nonzero scalar multiplies both the cross-product magnitude and the denominator by the scalar’s absolute value. The point \(A\) may also be replaced by any other point on the same line, because that changes \(\mathbf{q}-\mathbf{a}\) only by a multiple of \(\mathbf{v}\), whose cross product with \(\mathbf{v}\) is zero.

A Common Cross-Product Mistake

It is important not to treat the cross product itself as the perpendicular displacement. In the first worked example, \(\mathbf{d}=\langle 3,-1,4\rangle\), \(\mathbf{v}=\langle 2,-1,2\rangle\), and \(\mathbf{d}\times\mathbf{v}=\langle 2,2,-1\rangle\). A mistaken calculation might claim that the perpendicular displacement is \[ \frac{\mathbf{d}\times\mathbf{v}}{\lVert\mathbf{v}\rVert} =\left\langle\frac{2}{3},\frac{2}{3},-\frac{1}{3}\right\rangle. \] That is incorrect: this vector is not the displacement from the closest point on the line to \(Q\). In fact, the actual perpendicular displacement is \[ \mathbf{d}_{\perp} =\mathbf{d}-\frac{5}{3}\mathbf{v} =\left\langle-\frac{1}{3},\frac{2}{3},\frac{2}{3}\right\rangle. \] The two vectors differ in direction, though both have length \(1\). The correct use of the cross product is to take its magnitude and divide by the direction-vector length; the formula gives a distance, not the perpendicular displacement vector.

A second error is to divide by \(\lVert\mathbf{v}\rVert^2\) instead of \(\lVert\mathbf{v}\rVert\). The cross-product magnitude represents area and equals the perpendicular distance times \(\lVert\mathbf{v}\rVert\), not times its square. Always use \(\lVert\mathbf{d}\times\mathbf{v}\rVert/\lVert\mathbf{v}\rVert\).

Using the Distance Formula Reliably

A careful calculation follows a short sequence. First identify a point on the line and a nonzero direction vector. Next subtract the line point’s coordinates from the point’s coordinates to form \(\mathbf{d}\). Then compute \(\mathbf{d}\times\mathbf{v}\), take its magnitude, and divide by \(\lVert\mathbf{v}\rVert\). Keep the order of the cross product consistent; reversing it changes the cross product’s sign, but not its magnitude.

This method is useful whenever the line is given by a point and direction, including when it is written parametrically. It measures the perpendicular separation from the point to the entire line, not the distance to an arbitrary point chosen on that line. If the point lies on the line, \(\mathbf{d}\) is parallel to \(\mathbf{v}\), so the cross product is zero and the distance is zero, as it should be.