Projection Onto a Line Given by Two Points

Introduction

A line in space is often specified by two points rather than by a unit direction vector. To project a point onto that line, first use the two points to construct a direction vector. Then project the displacement from a point on the line to the given point onto that direction.

The resulting parameter tells us where the projected point lies along the line. It can be positive, zero, or negative: a line extends in both directions, so the nearest point need not lie between the two points originally used to specify it.

Representing the Line and the Given Point

Let \(A\) and \(B\) be distinct points on a line, with position vectors \(\mathbf{a}\) and \(\mathbf{b}\). Their difference gives a direction vector for the line:

Since \(A\) and \(B\) are distinct, \(\mathbf{d}\ne\mathbf{0}\). Every point on the line can be represented using a real parameter \(t\):

At \(t=0\), this gives \(A\); at \(t=1\), it gives \(B\). Values of \(t\) outside the interval from 0 to 1 are also allowed because the line continues beyond both points.

Now let \(P\) be the point being projected, with position vector \(\mathbf{p}\). The displacement from \(A\) to \(P\) is

The projection we need is the component of \(\mathbf{v}\) in the direction of \(\mathbf{d}\). Because \(\mathbf{d}\) need not have length 1, its squared length appears in the denominator.

Why This Formula Works

The earlier result for projection onto a unit direction says that if \(\mathbf{u}\) is a unit vector, the projection of \(\mathbf{v}\) onto that direction is \((\mathbf{v}\cdot\mathbf{u})\mathbf{u}\). To use that result, normalize the line direction: \(\mathbf{u}=\mathbf{d}/\|\mathbf{d}\|\). Then the projected displacement from \(A\) is

This projected displacement is \(t_*\mathbf{d}\), so its coefficient is exactly the parameter \(t_*\) in the line equation. Adding that displacement to \(\mathbf{a}\) gives the projected point \(\mathbf{q}\).

The projection is also characterized by a perpendicularity check. The displacement from \(P\) to the projected point is \(\mathbf{q}-\mathbf{p}\). Since \(\mathbf{q}=\mathbf{a}+t_*\mathbf{d}\) and \(\mathbf{p}=\mathbf{a}+\mathbf{v}\),

The last equality follows by substituting the formula for \(t_*\). Thus the displacement from \(P\) to \(\mathbf{q}\) is perpendicular to the line’s direction. The unit-direction projection result identifies \(\mathbf{q}\) as the nearest point on the line; the calculation above gives a convenient check on the result.

Worked Examples

A Common Mistake: Treating the Direction as a Unit Vector

The formula for projection onto a unit direction has no squared-length denominator because a unit vector has squared length 1. When the direction vector comes from two arbitrary points, it is not generally a unit vector. Omitting the denominator can therefore give the wrong parameter.

For example, take \(A=(0,0,0)\), \(B=(2,1,0)\), and \(P=(5,0,2)\). Then \(\mathbf{d}=(2,1,0)\) and \(\mathbf{v}=(5,0,2)\). Their dot product is \(10\), but \(\mathbf{d}\cdot\mathbf{d}=5\). The correct parameter is \(t_*=10/5=2\), giving \(Q=(4,2,0)\).

If we wrongly use \(t=10\), we get the line point \((20,10,0)\). It is on the line, but it is not the projection. In fact, the displacement from \(P\) to this false point is

The perpendicularity check exposes the error. For the correct point, the displacement from \(P\) to \(Q\) is \((4,2,0)-(5,0,2)=(-1,2,-2)\), and

Always calculate both \(\mathbf{v}\cdot\mathbf{d}\) and \(\mathbf{d}\cdot\mathbf{d}\), even when the direction components appear simple.

A Reliable Calculation Sequence

Why the Parameter Matters

The parameter does more than provide a convenient way to write the answer. It tells how the projected point sits relative to the defining points: \(t_*=0\) gives \(A\), \(t_*=1\) gives \(B\), a value between 0 and 1 gives a point between them, and values outside that range give points beyond one of them.

These interpretations concern projection onto an infinite line. If the target is instead the segment from \(A\) to \(B\), the parameter must be restricted to the interval from 0 to 1; simply using the line projection may then produce a point outside the segment. Keep the geometric object in the question clear before interpreting the result.