Geometric areas and perpendicular vectors | Year 12

01 / LEARN

Understand the idea.

Area from sides

For a triangle, first form two side vectors from the same vertex, such as $\vec{A B}$ and $\vec{A C}$ . The magnitude of their cross product gives a parallelogram area. The triangle occupies half that parallelogram.

Reversing the cross-product order changes the normal's direction but not the area. A zero cross product means the side vectors are parallel: the three points are collinear, so the triangle is degenerate and has area zero.

Choose the requested normal

A nonzero cross product gives one normal direction to the plane spanned by its inputs. Divide by its magnitude to obtain a unit normal. Unless an orientation is specified, the opposite unit normal is equally valid.

For a required normal length $k$ , multiply either unit normal by $k$ . Check both perpendicular dot products and the magnitude. Report areas in square units; a normal vector's scale is separate from the geometric area calculation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find the area of triangle A(0,0,0), B(2,0,0), C(0,3,1).

Worked solution

1
Form sides from A.
$\vec{A B} = (2, 0, 0), \vec{A C} = (0, 3, 1)$
Both displacements start at the same triangle vertex.
2
Calculate their cross product.
$(2, 0, 0) \times (0, 3, 1) = (0, - 2, 6)$
The ordered component formula supplies a vector normal to the triangle.
3
Find the parallelogram area.
$∣ (0, - 2, 6) ∣ = \sqrt{0^{2} + {(- 2)}^{2} + 6^{2}} = 2 \sqrt{10}$
The cross-product magnitude is the area spanned by the two sides.
4
Halve for the triangle.
$A = \frac{1}{2} (2 \sqrt{10}) = \sqrt{10} square units$
A diagonal divides the parallelogram into two equal-area triangles.

Answer

$\sqrt{10}$ square units

EXAMPLE 02

Worked example 2

Find both unit normals to (1,0,1) and (0,1,1).

Worked solution

1
Find one perpendicular direction.
$(1, 0, 1) \times (0, 1, 1) = (- 1, - 1, 1)$
A cross product is perpendicular to both input vectors.
2
Calculate its magnitude.
$∣ (- 1, - 1, 1) ∣ = \sqrt{1 + 1 + 1} = \sqrt{3}$
The three squared components sum to three.
3
Normalise and include both orientations.
$\hat{n} = \pm \frac{1}{\sqrt{3}} (- 1, - 1, 1)$
Dividing by the magnitude gives length one; neither orientation was prescribed.
4
Check the normal direction.
$(- 1, - 1, 1) \cdot (1, 0, 1) = 0, (- 1, - 1, 1) \cdot (0, 1, 1) = 0$
Scaling this direction by either unit-normal coefficient preserves perpendicularity.

Answer

$\pm (- 1, - 1, 1) / \sqrt{3}$

03 / YOUR TURN

Try it yourself.

Check your understanding

Do O, (1,2,3), (2,4,6) form a non-degenerate triangle?

Show answer & reasoning

No.

The vectors from $O$ are $(1, 2, 3)$ and $2 (1, 2, 3)$ . They are parallel, so their cross product and triangle area are zero; the triangle is degenerate.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the parallelogram area spanned by $(1, 0, 0)$ and $(0, 2, 2)$ .
2 marks
Find triangle area for $A (1, 0, 0)$ , $B (3, 0, 0)$ , $C (1, 4, 0)$ .
2 marks
Give a length-5 vector perpendicular to $(1, 0, 0)$ and $(0, 1, 0)$ .
2 marks
Find triangle area for sides $(1, 2, 0)$ and $(0, 1, 3)$ .
2 marks
Explain why replacing $b$ by $3 b$ triples the spanned parallelogram area.
2 marks
Find a unit normal to sides $(2, 0, 0)$ , $(0, 3, 4)$ with positive $z$ component.
2 marks

Read the worked solutions

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

Find the parallelogram area spanned by $(1, 0, 0)$ and $(0, 2, 2)$ .
2 marks
The cross product is $(0, - 2, 2)$ .
Its magnitude is $\sqrt{8}$ .
$2 \sqrt{2}$ square units
Find triangle area for $A (1, 0, 0)$ , $B (3, 0, 0)$ , $C (1, 4, 0)$ .
2 marks
The perpendicular sides have lengths 2 and 4.
Use half their product.
$4$ square units
Give a length-5 vector perpendicular to $(1, 0, 0)$ and $(0, 1, 0)$ .
2 marks
The cross-product direction is the $z$ axis.
Scale its unit direction by 5.
$(0, 0, 5)$ or $(0, 0, - 5)$
Find triangle area for sides $(1, 2, 0)$ and $(0, 1, 3)$ .
2 marks
Their cross product is $(6, - 3, 1)$ .
Its squared magnitude is 46; halve the magnitude.
$\frac{\sqrt{46}}{2}$ square units
Explain why replacing $b$ by $3 b$ triples the spanned parallelogram area.
2 marks
Use linearity of the cross product.
Use $∣ 3 v ∣ = 3 ∣ v ∣$ .
The cross product becomes $3 (a \times b)$ , so its magnitude triples.
Find a unit normal to sides $(2, 0, 0)$ , $(0, 3, 4)$ with positive $z$ component.
2 marks
The cross product is $(0, - 8, 6)$ with length 10.
Normalise using the positive- $z$ orientation.
$(0, - \frac{4}{5}, \frac{3}{5})$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

In this lesson

Understand the idea.

Area from sides

Choose the requested normal

See the method in action.

Worked example 1

Form sides from A.

Calculate their cross product.

Find the parallelogram area.

Halve for the triangle.

Worked example 2

Find one perpendicular direction.

Calculate its magnitude.

Normalise and include both orientations.

Check the normal direction.

Try it yourself.

Check your understanding

Keep practising