Finding a Perpendicular Foot and Distance to a Line

From a Point to Its Nearest Point on a Line

A line in three-dimensional space extends indefinitely in two directions. Given a point that may not lie on the line, we can ask which point on the line is closest to it. The answer is the perpendicular foot: the point where the perpendicular from the given point meets the line.

This construction turns a geometric question into a calculation with vectors. The line’s direction determines which point on it makes the connecting vector perpendicular to the line, and the length of that connecting vector is the distance we seek.

Line, Foot, and Distance

Suppose a line \(L\) passes through a point \(A\) and has a nonzero direction vector \(\mathbf{v}\). Every point on the line has the form \(A+t\mathbf{v}\), where \(t\) is a real number. Let \(P\) be the given point. We seek the point \(H\) on \(L\) for which the vector from \(H\) to \(P\) is perpendicular to the line’s direction.

Perpendicularity can be tested with a dot product: two vectors are perpendicular exactly when their dot product is zero. Since \(H=A+t\mathbf{v}\), the foot must satisfy \((P-H)\cdot\mathbf{v}=0\). Solving this condition gives the formula below.

The denominator is nonzero because \(\mathbf{v}\ne\mathbf{0}\), so \(\mathbf{v}\cdot\mathbf{v}=\|\mathbf{v}\|^2>0\). The vector \(P-A\) goes from the specified point on the line to the given point. Its dot product with the line direction determines how far along the line the foot lies.

Why the Formula Gives the Perpendicular Foot

Write any candidate point on the line as \(A+t\mathbf{v}\). If it is the foot, the connecting vector \(P-(A+t\mathbf{v})\) must be perpendicular to \(\mathbf{v}\). Thus its dot product with \(\mathbf{v}\) must equal zero. Expanding and solving gives the unique possible parameter:

This establishes that the formula finds a point on the line whose connecting vector is perpendicular to the line. It also explains why this point is the nearest one. If \(H=A+t_0\mathbf{v}\) is the calculated foot and \(X=A+s\mathbf{v}\) is any other point on the line, then \(X-H=(s-t_0)\mathbf{v}\). The vectors \(P-H\) and \(X-H\) are perpendicular. Since \(P-X=(P-H)-(X-H)\), the Pythagorean identity gives:

Lengths are nonnegative, so this inequality shows \(\|P-X\|\geq\|P-H\|\) for every point \(X\) on the line. Equality holds only when \(X=H\). Therefore, the perpendicular foot is the unique point on the line that minimizes the distance to \(P\).

Worked Examples

Common Error: Omitting the Denominator

A frequent mistake is to use \((P-A)\cdot\mathbf{v}\) directly as the line parameter. That shortcut works only when \(\mathbf{v}\cdot\mathbf{v}=1\), meaning that the direction vector is a unit vector. For a general direction vector, the denominator is essential.

In the first example, the numerator is \(12\), but \(\mathbf{v}\cdot\mathbf{v}=9\). If we incorrectly set \(t=12\), the proposed point is:

The vector from this incorrect point to \(P\) is \((-21,-8,23)\), and:

So the proposed point does not give a perpendicular connection. The correct parameter includes the denominator: \(t_0=12/9=4/3\). Always calculate both dot products before finding the foot.

A Reliable Calculation Sequence

To avoid mixing up points, vectors, and lengths, use the following order:

Interpreting the Result

The parameter \(t_0\) tells where the foot falls relative to the chosen point \(A\) and direction \(\mathbf{v}\). It need not be between zero and one: the line continues beyond \(A\) in both directions. A negative parameter places the foot in the direction opposite to \(\mathbf{v}\), while a positive parameter places it along \(\mathbf{v}\). The foot and distance do not depend on choosing a unit direction vector; the formula accounts for the length of any nonzero direction vector through its denominator.

Keep the two final quantities distinct. The foot \(H\) is a point on the line, while the distance is the nonnegative scalar \(\|P-H\|\). The vector \(P-H\) points from the foot to the given point; its length, not the vector itself, is the distance.