F6 - Perpendicular distances

Point-line distance by projection or cross product

A point is not on a line. Which point on the line is closest to it? Choosing the line's stated starting point rarely gives the shortest route. You need the foot of a perpendicular.

For a concrete case, let $A=(1,0,1)$, let the line have direction $\mathbf{u}=(1,1,0)$, and take $P=(3,0,1)$. Moving from $A$ along the line first brings you closer to $P$, then takes you away again. At the closest point $H$, the joining vector $\overrightarrow{HP}$ has zero scalar product with $\mathbf{u}$.

Write the line as $\mathbf{r}=\mathbf{a}+t\mathbf{u}$, where $\mathbf{u}\ne\mathbf{0}$, and put $\mathbf{v}=\mathbf{p}-\mathbf{a}=\overrightarrow{AP}$. If $H$ has parameter $t$, then $\overrightarrow{HP}=\mathbf{v}-t\mathbf{u}$. Perpendicularity gives

Since $\mathbf{u}\ne\mathbf{0}$, the denominator is positive. Once $H$ is found, the distance is $|\overrightarrow{HP}|$. Any other point $Q$ on the line makes a right triangle with $P$ and $H$, so $PQ^2=PH^2+HQ^2\ge PH^2$. This proves the perpendicular route is shortest, including distance zero when $P$ lies on the line.

There is a second route when only the distance is wanted. The parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ has area $|\mathbf{v}\times\mathbf{u}|$; it also has base $|\mathbf{u}|$ and height equal to the required distance.

Use projection when the foot itself is required; the cross product is often shorter when only distance is required.

Keep the line fixed and change only $P$'s height from $1$ to $4$. The line direction has zero third component, so the projection parameter will stay the same.

In the cross-section below, both the line and the point lie in the page. To construct your own distance diagram, draw the line, mark a point $A$ on it and place $P$ off it. Drop a perpendicular from $P$ to the line, label its foot $H$, and mark the right angle. The triangle $AHP$ separates movement along the line from the perpendicular distance; its page is the plane containing the line and $P$, rather than necessarily a coordinate plane.

[DIAGRAM: asset_name: A perpendicular foot gives the shortest route; asset_slug: aqa7367-point-line-foot; caption: The displacement AP splits into a component along the line and the perpendicular HP; description: An equal-scale cross-section through A, H and P from Worked example 2. A is drawn at (0,0), H at (square root of two,0) and P at (square root of two,square root of eleven) in local orthogonal coordinates. The horizontal blue line extends through A and H. A dashed grey AP has length square root of thirteen, the blue AH has length square root of two and the clay perpendicular HP has length square root of eleven. A right-angle marker appears at H. The actual three-dimensional coordinates A equals (1,0,1), H equals (2,1,1) and P equals (3,0,4) are labelled separately.]

The diagram also explains the area formula: the same height is perpendicular to every base segment along the line. Reading off $AP$ instead would use the hypotenuse and overestimate the distance.

Find both the foot and distance for a fresh line.

Point-plane distance along a normal

To reach a plane by the shortest route, move along its normal. Movement within the plane can only make a joining segment longer.

For example, the distance from $(2,1,1)$ to $x+2y+2z=1$ is not the equation's residual $2+2+2-1=5$. That number uses the normal $(1,2,2)$, whose length is $3$. Dividing by the normal length converts the residual into a distance.

For $\mathbf{n}\cdot\mathbf{r}=d$, with $\mathbf{n}\ne\mathbf{0}$, choose any point $A$ in the plane. The scalar projection of $\overrightarrow{AP}$ onto the normal direction is

It is signed, indicating which side of the plane contains $P$. Its absolute value is the distance. To find the foot, subtract the vector projection onto the normal:

This point lies in the plane because taking its dot product with $\mathbf{n}$ gives $d$; its displacement from $P$ is a normal multiple. These two checks establish that it is the perpendicular foot.

If the plane is given as $ax+by+cz+e=0$, the numerator is $|ax_0+by_0+cz_0+e|$. Keep the equation's sign convention consistent. A point already in the plane gives distance zero.

Reflect the point through $A=(1,0,0)$, which lies in the plane. This changes $P$ to $(0,-1,-1)$, reversing the residual but preserving the distance.

Multiplying the entire plane equation by any nonzero constant changes both numerator and denominator by the same absolute factor, so it cannot change the distance.

Use the formula with a new plane and a new point.

Distances between skew or parallel lines

For two skew lines, the shortest joining segment must be perpendicular to both directions. For two parallel lines, a joining segment perpendicular to their shared direction gives the shortest route.