What is the foot of the perpendicular to a line?

The foot F of the perpendicular from a point P to a line r = a + λd is the point on the line closest to P. Parametrise F = a + λd, then impose

What is the foot of the perpendicular to a plane?

The foot F of the perpendicular from P to a plane lies along the normal n through P. Write the line r = p + tn and substitute into the plane equation to find t, then F. Equivalently, step from P along n by the signed distance.

What is foot to a line without the perpendicularity condition?

The defining condition is PF·d = 0; guessing F without it gives the wrong point.

What is ratio theorem the wrong way round?

The point dividing AB in ratio m : n weights b by m and a by n (the opposite of the naive guess); check with an endpoint.

SingaporeFurther MathsSyllabus dot point

How do we use vector methods to prove geometric results and find reflections, foot of perpendicular and other constructions?

Apply vector methods to geometric problems including the foot of the perpendicular, reflections of points, and proofs of geometric properties

A focused answer to the H2 Further Mathematics outcome on applying vectors to geometry. Finding the foot of the perpendicular from a point to a line or plane, reflecting a point in a line or plane, and using position vectors and the ratio theorem to prove geometric results.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

SEAB wants you to apply vector methods to geometric problems: finding the foot of the perpendicular from a point to a line or plane, reflecting a point in a line or plane, and proving geometric properties using position vectors. These tie together the scalar and vector products with the equations of lines and planes.

The answer

The foot of the perpendicular to a line

The foot $F$ of the perpendicular from a point $P$ to a line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ is the point on the line closest to $P$ . Parametrise $F = \mathbf{a} + \lambda\mathbf{d}$ , then impose

\overrightarrow{PF}\cdot\mathbf{d} = 0,

since the shortest connection is perpendicular to the line. Solve for $\lambda$ and substitute back for $F$ .

The foot of the perpendicular to a plane

The foot $F$ of the perpendicular from $P$ to a plane lies along the normal $\mathbf{n}$ through $P$ . Write the line $\mathbf{r} = \mathbf{p} + t\mathbf{n}$ and substitute into the plane equation to find $t$ , then $F$ . Equivalently, step from $P$ along $\mathbf{n}$ by the signed distance.

Reflecting a point

The reflection $P'$ of $P$ in a line or plane is on the far side of the mirror, the same distance away. Once the foot $F$ is known, the foot is the midpoint of $P$ and $P'$ , so

P' = 2F - P.

This single relation handles reflection in both a line and a plane.

Proving geometric results with vectors

Position vectors turn geometry into algebra. Useful tools: the midpoint of $A$ and $B$ is $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ ; the point dividing $AB$ in ratio $m : n$ is $\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$ (the ratio theorem); two segments are parallel when their vectors are scalar multiples, and three points are collinear when two of the joining vectors are parallel. Showing such relations proves results like "the diagonals of a parallelogram bisect each other".

Worked example

Prove, using vectors, that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Step 1: Set up position vectors

Let the triangle have vertices $A$ , $B$ , $C$ with position vectors $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ . Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$ .

Step 2: Write the midpoints

\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}), \qquad \mathbf{n} = \tfrac{1}{2}(\mathbf{a} + \mathbf{c}).

Step 3: Form the joining vector

\overrightarrow{MN} = \mathbf{n} - \mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{c}) - \tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \tfrac{1}{2}(\mathbf{c} - \mathbf{b}).

Step 4: Compare with the third side

The third side is $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$ . So $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$ .

Step 5: Conclude

Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{BC}$ , the segment $MN$ is parallel to $BC$ ; and the factor $\tfrac{1}{2}$ shows it is half the length. This proves the midpoint theorem.

Common traps

Reflecting with the wrong relation: The reflection is $P' = 2F - P$ , not $F - P$ or $2P - F$ ; the foot is the midpoint of $P$ and $P'$ .
Foot to a line without the perpendicularity condition: The defining condition is $\overrightarrow{PF}\cdot\mathbf{d} = 0$ ; guessing $F$ without it gives the wrong point.
Ratio theorem the wrong way round: The point dividing $AB$ in ratio $m : n$ weights $\mathbf{b}$ by $m$ and $\mathbf{a}$ by $n$ (the opposite of the naive guess); check with an endpoint.
Assuming collinearity without proof: To prove three points collinear, show two joining vectors are parallel (scalar multiples) and share a point.
Mixing up the normal and direction: Foot to a plane moves along the normal $\mathbf{n}$ ; foot to a line uses the line's direction $\mathbf{d}$ . Use the right vector for the object.

Examples in context

Example 1. Mirror image in graphics. Rendering a reflection in a flat mirror computes $P' = 2F - P$ for each visible point, where $F$ is its foot on the mirror plane; the whole reflected scene is built from this one vector construction.

Example 2. Proving a quadrilateral is a parallelogram. Showing that the midpoints of the sides of any quadrilateral form a parallelogram (Varignon's theorem) is a pure position-vector argument, demonstrating how vector algebra replaces lengthy classical geometry proofs.

Try this

Q1. State the condition that determines the foot of the perpendicular from $P$ to a line with direction $\mathbf{d}$ . [1 mark]

Cue. $\overrightarrow{PF}\cdot\mathbf{d} = 0$ , with $F$ a point on the line.

Q2. If $F$ is the foot of the perpendicular from $P$ to a plane, write the reflection $P'$ . [1 mark]

Cue. $P' = 2F - P$ .

Q3. Write the position vector of the midpoint of points with position vectors $\mathbf{a}$ and $\mathbf{b}$ . [1 mark]

Cue. $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksFind the foot of the perpendicular from the point

P(4, 1, 2)

to the line

\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}

Show worked answer →

A general point on the line is $F = (\lambda, \lambda, \lambda)$ . The foot of perpendicular $F$ satisfies $\overrightarrow{PF}\perp\mathbf{d}$ , where $\mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ .

$\overrightarrow{PF} = F - P = \begin{pmatrix} \lambda - 4 \\ \lambda - 1 \\ \lambda - 2 \end{pmatrix}$ . Require $\overrightarrow{PF}\cdot\mathbf{d} = 0$ :

(\lambda - 4) + (\lambda - 1) + (\lambda - 2) = 0 \Rightarrow 3\lambda - 7 = 0 \Rightarrow \lambda = \tfrac{7}{3}.

F = \left(\tfrac{7}{3}, \tfrac{7}{3}, \tfrac{7}{3}\right)

Markers reward parametrising $F$ on the line, the perpendicularity condition $\overrightarrow{PF}\cdot\mathbf{d} = 0$ , solving for $\lambda = \tfrac{7}{3}$ , and the foot $\left(\tfrac{7}{3}, \tfrac{7}{3}, \tfrac{7}{3}\right)$ .

Original6 marksThe point

P

has position vector

(3, 0, 4)

. Find the reflection

P'

P

in the plane

x + 2y + 2z = 0

Show worked answer →

The reflection lies along the normal $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ through $P$ . The foot of perpendicular from $P$ to the plane is found by moving from $P$ along $\mathbf{n}$ by the signed distance.

Signed value: $\dfrac{\mathbf{p}\cdot\mathbf{n} - d}{|\mathbf{n}|^2} = \dfrac{(3 + 0 + 8) - 0}{1 + 4 + 4} = \dfrac{11}{9}$ .

The foot is $F = P - \dfrac{11}{9}\mathbf{n}$ , and the reflection is twice as far: $P' = P - 2\cdot\dfrac{11}{9}\mathbf{n} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} - \dfrac{22}{9}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ .

P' = \begin{pmatrix} 3 - \tfrac{22}{9} \\ 0 - \tfrac{44}{9} \\ 4 - \tfrac{44}{9} \end{pmatrix} = \begin{pmatrix} \tfrac{5}{9} \\ -\tfrac{44}{9} \\ -\tfrac{8}{9} \end{pmatrix}.

Markers reward moving along the normal, the signed multiple $\tfrac{11}{9}$ , doubling it for the reflection, and the reflected point $P' = \left(\tfrac{5}{9}, -\tfrac{44}{9}, -\tfrac{8}{9}\right)$ .