Cross product and determinant calculations | Year 12

01 / LEARN

Understand the idea.

Construct a perpendicular vector

The cross product produces a vector perpendicular to two given vectors. Compute its components in the cyclic order $(a_{y} b_{z} - a_{z} b_{y}, a_{z} b_{x} - a_{x} b_{z}, a_{x} b_{y} - a_{y} b_{x})$ . Keeping this order avoids the common middle-component sign error.

In a right-handed coordinate system, the right-hand rule fixes orientation. Reversing the input order negates the result: $b \times a = - a \times b$ . Parallel inputs give the zero vector, which cannot be normalised to a unit normal.

Check size and orientation

Check a calculated cross product by dotting it with each input. Both products should be zero. Its magnitude is $∣ a ∣ ∣ b ∣ sin θ$ , the area of their parallelogram; magnitude alone cannot distinguish the two opposite orientations.

A scalar triple product $a \cdot (b \times c)$ combines a normal direction with a third vector. Its absolute value is the parallelepiped volume. Its sign records orientation, while a zero result means the three vectors are coplanar.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find (1,2,0) cross (0,1,3).

Worked solution

1
Substitute in the component formula.
$\begin{aligned} c_{x} & = 2 \cdot 3 - 0 \cdot 1 = 6 \\ c_{y} & = 0 \cdot 0 - 1 \cdot 3 = - 3 \\ c_{z} & = 1 \cdot 1 - 2 \cdot 0 = 1 \end{aligned}$
Each component of the cross product uses one ordered difference of products.
2
Simplify all three components.
$a \times b = (6, - 3, 1)$
The middle difference is $0 - 3 = - 3$ .
3
Check perpendicularity to both inputs.
$\begin{aligned} (6, - 3, 1) \cdot (1, 2, 0) & = 6 - 6 = 0 \\ (6, - 3, 1) \cdot (0, 1, 3) & = - 3 + 3 = 0 \end{aligned}$
Both zero dot products confirm perpendicularity; the ordered formula fixes its sign.

Answer

$(6, - 3, 1)$

EXAMPLE 02

Worked example 2

Find $(1, 0, 0) \cdot [(0, 2, 0) \times (0, 0, 3)]$ .

Worked solution

1
Calculate the inner cross product.
$(0, 2, 0) \times (0, 0, 3) = (2 \cdot 3, 0, 0) = (6, 0, 0)$
The positive $y$ direction crossed with the positive $z$ direction gives positive $x$ .
2
Take the dot product with the remaining vector.
$(1, 0, 0) \cdot (6, 0, 0) = 6$
The outer dot product gives a scalar.
3
Interpret its sign and magnitude.
$V = ∣ 6 ∣ = 6$
The triple product is positive for this ordering, and its absolute value is the spanned volume.

Answer

$6$

03 / YOUR TURN

Try it yourself.

Check your understanding

What is (2,4,6) cross (1,2,3)?

Show answer & reasoning

The zero vector.

Since $(2, 4, 6) = 2 (1, 2, 3)$ , the vectors are parallel. Their angle has zero sine, so their cross product is the zero vector.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $(1, 0, 1) \times (0, 2, 1)$ .
2 marks
Find $(0, 2, 1) \times (1, 0, 1)$ .
2 marks
Find $i \times j$ and $j \times i$ .
2 marks
Is $(3, 0, - 1)$ perpendicular to $(1, 2, 3) \times (2, - 1, 0)$ ?
2 marks
Find the determinant with rows $(1, 2, 0)$ , $(0, 1, 3)$ , $(2, 0, 1)$ .
2 marks
Find the volume spanned by $(2, 0, 0)$ , $(0, 3, 0)$ , $(0, 0, 4)$ .
2 marks

Read the worked solutions

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

Find $(1, 0, 1) \times (0, 2, 1)$ .
2 marks
Use $(0 \cdot 1 - 1 \cdot 2, 1 \cdot 0 - 1 \cdot 1, 1 \cdot 2 - 0 \cdot 0)$ .
Check both input dot products are zero.
$(- 2, - 1, 2)$
Find $(0, 2, 1) \times (1, 0, 1)$ .
2 marks
Reverse the order of the previous cross product.
Negate every component.
$(2, 1, - 2)$
Find $i \times j$ and $j \times i$ .
2 marks
The positive basis is right-handed.
Reversing the order negates the result.
$k$ and $- k$
Is $(3, 0, - 1)$ perpendicular to $(1, 2, 3) \times (2, - 1, 0)$ ?
2 marks
The cross product is $(3, 6, - 5)$ .
Its dot product with $(3, 0, - 1)$ is 14.
No.
Find the determinant with rows $(1, 2, 0)$ , $(0, 1, 3)$ , $(2, 0, 1)$ .
2 marks
Cross the last two rows: $(1, 6, - 2)$ .
Dot with the first: $1 + 12 = 13$ .
$13$
Find the volume spanned by $(2, 0, 0)$ , $(0, 3, 0)$ , $(0, 0, 4)$ .
2 marks
The determinant is $2 \cdot 3 \cdot 4 = 24$ .
Take its absolute value for volume.
$24$ cubic units

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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.

Construct a perpendicular vector

Check size and orientation

See the method in action.

Worked example 1

Substitute in the component formula.

Simplify all three components.

Check perpendicularity to both inputs.

Worked example 2

Calculate the inner cross product.

Take the dot product with the remaining vector.

Interpret its sign and magnitude.

Try it yourself.

Check your understanding

Keep practising