Perpendicularity | Year 11

01 / LEARN

Understand the idea.

An algebraic right-angle test

For non-zero vectors, a zero dot product is exactly the condition for perpendicularity. The geometric formula reduces to $cos θ = 0$ , since both magnitudes are positive. This test handles horizontal and vertical directions without needing slopes.

A quarter-turn sends $(a, b)$ to $(- b, a)$ or $(b, - a)$ . Their dot products with $(a, b)$ are zero because the two products cancel. These give useful perpendicular directions whenever $(a, b)$ is non-zero.

(a, b) \cdot (- b, a) = 0

Equal-scale coordinate axes with u=(3,1) and a labelled perpendicular vector (-1,3), both drawn from the origin. Tick intervals are one unit.

Use the angle's actual vertex

For an angle named $A B C$ , the middle letter is its vertex. Use $\vec{B A}$ and $\vec{B C}$ , with both arrows starting at B. Form the displacements before using the dot product so the calculation tests the intended angle.

Check that both displacements are non-zero. A coincident pair of points gives a zero displacement, which has zero dot product with every vector but no direction. A zero calculation alone cannot turn a degenerate angle into a right angle.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Finding an unknown component

Find k if $(3, k)$ is perpendicular to $(4, - 2)$ .

Worked solution

1
Translate perpendicularity into an equation.
$(3, k) \cdot (4, - 2) = 0$
Both vectors are non-zero, so a zero dot product is equivalent to a right angle.
2
Expand and solve.
$\begin{aligned} 12 - 2 k & = 0, \\ 2 k & = 12, \\ k & = 6. \end{aligned}$
Matching component products give a linear equation for the unknown.
3
Verify the required condition.
$(3, 6) \cdot (4, - 2) = 12 - 12 = 0$
The resulting non-zero vector is perpendicular to the given direction.

Answer

$k = 6$ .

EXAMPLE 02

A right angle in a triangle

A=(1,1), B=(4,2), C=(3,5). Show that angle ABC is a right angle.

Worked solution

1
Form the two sides from the angle vertex B.
$\begin{aligned} \vec{B A} & = (1, 1) - (4, 2) = (- 3, - 1), \\ \vec{B C} & = (3, 5) - (4, 2) = (- 1, 3) . \end{aligned}$
Angle ABC is formed by the two arrows starting at B.
2
Calculate their dot product.
$\vec{B A} \cdot \vec{B C} = (- 3) (- 1) + (- 1) (3) = 3 - 3 = 0$
The matching component products cancel.
3
Conclude that the angle is right.
$∠ A B C = 90^{\circ}$
Both side vectors are non-zero, so their zero dot product proves perpendicularity.

Answer

$∠ A B C = 90^{\circ}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Give a non-zero vector perpendicular to $(5, - 2)$ .

Show answer & reasoning

For example $(2, 5)$ .

A quarter-turn of $(5, - 2)$ is $(2, 5)$ . It is non-zero and $(5, - 2) \cdot (2, 5) = 10 - 10 = 0$ , so it is perpendicular. Any non-zero scalar multiple of it also works.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Are $(2, 3)$ and $(6, - 4)$ perpendicular?
2 marks
Find $k$ if $(k, 5)$ is perpendicular to $(2, - 6)$ .
2 marks
Find a unit vector perpendicular to $(3, 4)$ .
2 marks
$A = (0, 0)$ , $B = (2, 1)$ , $C = (1, 3)$ . Decide whether angle $A B C$ is a right angle.
3 marks
For $u = (t, 2)$ , $v = (3, t)$ , find all real $t$ making $u$ perpendicular to $v$ .
3 marks
Explain why $0 \cdot (1, 2) = 0$ does not show a right angle.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Are $(2, 3)$ and $(6, - 4)$ perpendicular?
2 marks
Their dot product is $12 - 12 = 0$ .
Yes.
Find $k$ if $(k, 5)$ is perpendicular to $(2, - 6)$ .
2 marks
$2 k - 30 = 0$ .
$k = 15$ .
Find a unit vector perpendicular to $(3, 4)$ .
2 marks
Rotate to $(- 4, 3)$ and divide by its length five.
For example $(- \frac{4}{5}, \frac{3}{5})$ .
$A = (0, 0)$ , $B = (2, 1)$ , $C = (1, 3)$ . Decide whether angle $A B C$ is a right angle.
3 marks
$B A = (- 2, - 1)$ ; $B C = (- 1, 2)$ .
Their dot product is zero.
Yes.
For $u = (t, 2)$ , $v = (3, t)$ , find all real $t$ making $u$ perpendicular to $v$ .
3 marks
$u \cdot v = 3 t + 2 t = 5 t$ .
Set it equal to zero; the resulting vectors are both non-zero.
$t = 0$ .
Explain why $0 \cdot (1, 2) = 0$ does not show a right angle.
2 marks
The geometric perpendicularity conclusion requires both vectors to be non-zero.
The zero vector has no direction, so it forms no defined angle.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: vectors in the plane. Covers magnitude, components, geometry, scalar products, projections and proofs; follows the approved Specialist preparation structure.

Official curriculum source ↗

In this lesson

Understand the idea.

An algebraic right-angle test

Use the angle's actual vertex

See the method in action.

Finding an unknown component

Translate perpendicularity into an equation.

Expand and solve.

Verify the required condition.

A right angle in a triangle

Form the two sides from the angle vertex B.

Calculate their dot product.

Conclude that the angle is right.

Try it yourself.

Check your understanding

Keep practising