Foot of the Perpendicular and Shortest Distances

The closest point on a line to a given point is where a perpendicular meets it. Find it with one scalar product, then get distances, reflections and areas almost for free.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14AQA 7367 Paper 1≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find the foot of the perpendicular from a point to a line using $\overrightarrow{AP}\cdot\mathbf{d}=0$ .
Find the perpendicular (shortest) distance from a point to a line.
Reflect a point in a line.
Use projections and Pythagoras as a check, and to find areas.

Pause and think

You are standing in a field next to a long straight road. Which point on the road is closest to you, and how could you recognise it?

Section 2 of 14

Core idea

Start hereThe closest approach

A ship sails a straight course past a lighthouse. When is it closest, and how close? At the closest moment, the line from the lighthouse to the ship is perpendicular to the course. The scalar product turns that geometric fact into an equation.

The big idea

For point $P$ and line $\mathbf r=\mathbf a+\lambda\mathbf d$ : write a general point $Q(\lambda)$ , set $\overrightarrow{PQ}\cdot\mathbf d=0$ , solve for $\lambda$ to find the foot $F$ ; the shortest distance is $|\overrightarrow{PF}|$ .

1. Why perpendicular is shortest

Distance from P to a moving point Q on the line

λ = -0.60

Q

slides along the line, the dashed distance

PQ

is smallest exactly when

Q

reaches

F

, where

PF

meets the line at right angles.

Pythagoras argument

$F\text{ with }PF\perp\text{line};\ Q\text{ any other point on the line}$
Setup.
$PQ^2=PF^2+FQ^2$
Right angle at $F$ .
$PQ^2>PF^2\ (FQ\neq0)$
So $PF$ is the shortest.

2. The method in 3D

Worked example $P(5,4,2)$ ; line $\mathbf r=(1,0,2)+\lambda(1,2,-1)$

Find the foot of the perpendicular and the distance.

Goal: $F$ and $|PF|$ .
Strategy: General point; dot product zero.

$Q=(1+\lambda,\ 2\lambda,\ 2-\lambda)$
General point.
$\overrightarrow{PQ}=(\lambda-4,\ 2\lambda-4,\ -\lambda)$
$Q-P$ .
$(\lambda-4)+2(2\lambda-4)-(-\lambda)=0\Rightarrow6\lambda-12=0$
$\overrightarrow{PQ}\cdot\mathbf d=0$ .
$\lambda=2:\ F=(3,4,0),\ \overrightarrow{PF}=(-2,0,-2)$
Foot.
$|PF|=\sqrt8=2\sqrt2$
Distance.

Answer

F(3,4,0)

; distance

2\sqrt2

✓ Check

\overrightarrow{PF}\cdot\mathbf d=-2+0+2=0

✓.

TrickReflection of P in the line

P'=F+\overrightarrow{PF}=2F-P=(1,4,-2)

Pause and think

Distance from the origin to the line $\mathbf r=(0,3,0)+\lambda(1,0,0)$ ?

Summary

General point $Q(\lambda)$ .
$\overrightarrow{PQ}\cdot\mathbf d=0$ .
Distance $|PF|$ .

Foot of perpendicular: Closest point on the line.

HistoryThe method of least squares. Around 1795 the teenage Carl Friedrich Gauss developed the method of least squares: choose the line that makes the total of the squared distances to the data points as small as possible. In 1801 he used it to predict where the lost dwarf planet Ceres would reappear, and astronomers found it almost exactly where he said. Finding shortest distances is at the heart of the method.

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