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Cross product and applications: VCE Specialist Mathematics Unit 3

Syllabus dot point

“The vector (cross) product of two three-dimensional vectors, its definition in component form, the geometric meaning of its direction and magnitude, and its applications to finding a normal vector, the area of a parallelogram or triangle, and testing for parallel vectors”

6 min read

Quick answer

The cross product a×b\mathbf{a} \times \mathbf{b} of two vectors in three dimensions is a vector perpendicular to both, with direction given by the right-hand rule and magnitude ∣a∣∣b∣sin⁡θ|\mathbf{a}||\mathbf{b}|\sin\theta, where θ\theta is the angle between them. In components, a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}. Its magnitude equals the area of the parallelogram spanned by a\mathbf{a} and b\mathbf{b}, so the triangle area is half that. The cross product is the zero vector exactly when the vectors are parallel, giving a parallel test.

Jump to a section
  1. What this dot point is asking
  2. Definition in components
  3. Direction and magnitude
  4. Applications
  5. Examples in context
  6. Try this
  7. Exam-style questions

What this dot point is asking

VCAA wants you to compute the cross product of two three-dimensional vectors, understand that it produces a vector perpendicular to both, interpret its magnitude as an area, and apply it to find normal vectors, areas of parallelograms and triangles, and to test whether two vectors are parallel. The cross product underpins the equation of a plane studied alongside it.

Definition in components

For a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} and b=b1i+b2j+b3k\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}, the cross product is

a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k.\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}.

A reliable way to compute it is the determinant expansion

a×b=∣ijka1a2a3b1b2b3∣,\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix},

expanding along the top row and remembering the middle term carries a minus sign.

Direction and magnitude

The vector a×b\mathbf{a} \times \mathbf{b} is perpendicular to both a\mathbf{a} and b\mathbf{b}, so it is normal to the plane they span. Its direction follows the right-hand rule: point the fingers from a\mathbf{a} toward b\mathbf{b} and the thumb gives the direction of the product. Reversing the order reverses the direction, so b×a=−(a×b)\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}); the cross product is not commutative.

Its magnitude is

∣a×b∣=∣a∣ ∣b∣sin⁡θ,|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta,

which equals the area of the parallelogram with sides a\mathbf{a} and b\mathbf{b}.

Applications

Normal vector
Since a×b\mathbf{a} \times \mathbf{b} is perpendicular to both, it provides a normal to the plane containing them, which is exactly what is needed to write the plane's equation.
Area
The parallelogram on a\mathbf{a} and b\mathbf{b} has area ∣a×b∣|\mathbf{a} \times \mathbf{b}|, and the triangle with those two sides has area 12∣a×b∣\frac{1}{2}|\mathbf{a} \times \mathbf{b}|.
Parallel test
If a\mathbf{a} and b\mathbf{b} are parallel, θ=0\theta = 0 or π\pi, so sin⁡θ=0\sin\theta = 0 and a×b=0\mathbf{a} \times \mathbf{b} = \mathbf{0}. Conversely a zero cross product (for non-zero vectors) means they are parallel.
Worked example

Area of a triangle in space

Find the area of the triangle with vertices A(1,0,2)A(1, 0, 2), B(3,1,1)B(3, 1, 1) and C(2,4,3)C(2, 4, 3).

Step 1: Form two edge vectors from AA

Choosing one vertex as the base and subtracting its coordinates from the others gives two vectors that define the triangle. These become the inputs to the cross product.

AB→=(3−1, 1−0, 1−2)=(2,1,−1),AC→=(2−1, 4−0, 3−2)=(1,4,1).\overrightarrow{AB} = (3 - 1,\ 1 - 0,\ 1 - 2) = (2, 1, -1), \qquad \overrightarrow{AC} = (2 - 1,\ 4 - 0,\ 3 - 2) = (1, 4, 1).

Step 2: Compute the cross product using the determinant

The cross product gives a vector perpendicular to both edges. The middle (j\mathbf{j}) component carries a minus sign in the determinant expansion, which is the most common slip.

AB→×AC→=∣ijk21−1141∣.\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & 4 & 1 \end{vmatrix}.

  • i\mathbf{i} component: (1)(1)−(−1)(4)=1+4=5(1)(1) - (-1)(4) = 1 + 4 = 5.
  • j\mathbf{j} component: −[(2)(1)−(−1)(1)]=−(2+1)=−3-\big[(2)(1) - (-1)(1)\big] = -(2 + 1) = -3.
  • k\mathbf{k} component: (2)(4)−(1)(1)=8−1=7(2)(4) - (1)(1) = 8 - 1 = 7.

So AB→×AC→=(5,−3,7)\overrightarrow{AB} \times \overrightarrow{AC} = (5, -3, 7).

Step 3: Find the magnitude and halve it for the triangle area

The magnitude of the cross product equals the area of the parallelogram spanned by the two edge vectors. The triangle occupies exactly half that parallelogram.

∣(5,−3,7)∣=25+9+49=83.|(5, -3, 7)| = \sqrt{25 + 9 + 49} = \sqrt{83}.

Area=1283≈4.56.\text{Area} = \frac{1}{2}\sqrt{83} \approx 4.56.

Check the perpendicularity: AB→⋅(5,−3,7)=10−3−7=0\overrightarrow{AB} \cdot (5, -3, 7) = 10 - 3 - 7 = 0, confirming the product is normal to AB→\overrightarrow{AB} as expected.

Common exam traps
Dropping the middle minus sign
The j\mathbf{j} component of the determinant expansion is subtracted. Forgetting this flips its sign.
Confusing cross with dot
The dot product gives a scalar; the cross product gives a vector. The area formula uses sin⁡θ\sin\theta (cross), not cos⁡θ\cos\theta (dot).
Forgetting the half for a triangle
The cross-product magnitude is the parallelogram area; halve it for a triangle.
Order matters
a×b=−b×a\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}, so swapping the vectors reverses the normal's direction.

Examples in context

Example 1. i×j=k\mathbf{i} \times \mathbf{j} = \mathbf{k}, illustrating the right-hand rule for the standard basis.

Example 2. If a×b=0\mathbf{a} \times \mathbf{b} = \mathbf{0} for non-zero a,b\mathbf{a}, \mathbf{b}, the vectors are parallel, so one is a scalar multiple of the other.

Try this

Q1. Compute (1,0,0)×(0,1,0)(1, 0, 0) \times (0, 1, 0). [1 mark]

  • Cue. (0,0,1)(0, 0, 1).

Q2. Find the area of the parallelogram on (2,0,0)(2, 0, 0) and (0,3,0)(0, 3, 0). [2 marks]

  • Cue. Cross product (0,0,6)(0, 0, 6), magnitude 66.

Q3. Show (2,4,6)(2, 4, 6) and (1,2,3)(1, 2, 3) are parallel using the cross product. [2 marks]

  • Cue. Cross product is 0\mathbf{0}, so they are parallel.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style1 mark
A plane contains the points A(1, 3, -2) and the point D(0, 2, 0), giving vectors AB = -2i - 5j + 6k and AD = -i - j + 2k. AB and AD are adjacent sides of a parallelogram. Find the area of this parallelogram.
Show worked answer →

The area of a parallelogram with adjacent sides AB and AD is the magnitude of their cross product, |AB x AD|.

Compute the cross product of AB = (-2, -5, 6) and AD = (-1, -1, 2):
i component: (-5)(2) - (6)(-1) = -10 + 6 = -4
j component: -[ (-2)(2) - (6)(-1) ] = -[ -4 + 6 ] = -2
k component: (-2)(-1) - (-5)(-1) = 2 - 5 = -3
So AB x AD = -4i - 2j - 3k.

Its magnitude is sqrt((-4)^2 + (-2)^2 + (-3)^2) = sqrt(16 + 4 + 9) = sqrt(29).

Therefore the area of the parallelogram is sqrt(29) square units.

2023 VCAA-style2 marks
The points with coordinates A(1, 1, 2), B(1, 2, 3) and C(3, 2, 4) all lie in a plane. Find the vectors AB and AC, and hence show that the area of triangle ABC is 1.5 square units.
Show worked answer →

Find the edge vectors from A: AB = B - A = (0, 1, 1) and AC = C - A = (2, 1, 2).

The area of triangle ABC is half the magnitude of the cross product AB x AC.

Compute AB x AC with AB = (0, 1, 1) and AC = (2, 1, 2):
i component: (1)(2) - (1)(1) = 2 - 1 = 1
j component: -[ (0)(2) - (1)(2) ] = -[ 0 - 2 ] = 2
k component: (0)(1) - (1)(2) = 0 - 2 = -2
So AB x AC = i + 2j - 2k, with magnitude sqrt(1^2 + 2^2 + (-2)^2) = sqrt(9) = 3.

Area of triangle = (1/2)|AB x AC| = (1/2)(3) = 1.5 square units, as required.

2025 VCAA-style2 marks
Consider planes P2: 3x + 6y + 5z = 7 and P3: x + 9y - 3z = 7. Find a vector that gives the direction of the line of intersection of the planes P2 and P3.
Show worked answer →

The line of intersection lies in both planes, so it is perpendicular to both normal vectors. A direction for the line is therefore the cross product of the two normals.

The normals are n2 = (3, 6, 5) and n3 = (1, 9, -3). Compute n2 x n3:
i component: (6)(-3) - (5)(9) = -18 - 45 = -63
j component: -[ (3)(-3) - (5)(1) ] = -[ -9 - 5 ] = 14
k component: (3)(9) - (6)(1) = 27 - 6 = 21
So n2 x n3 = -63i + 14j + 21k.

This is a valid direction vector. Dividing by the common factor 7 gives the simpler equivalent direction -9i + 2j + 3k.

Practise this

Sources & how we know this

ExamExplained