Cross product of vectors (HL only)

IB Mathematics: Analysis and Approaches HL· 3.15 Cross product of vectors· 15 min read

1. Definition and Geometric Interpretation★★☆☆☆HL only⏱ 5 min

Unlike the dot product (which returns a scalar), the cross product of two 3D vectors returns another 3D vector. This vector has a specific magnitude and direction defined relative to the two input vectors.

📘 Definition

Cross Product

$a \times b$

A vector perpendicular to both $a$ and $b$ with magnitude $∣ a ∣∣ b ∣ sin θ$ , where $θ$ (for $0 \leq θ \leq π$ ) is the angle between $a$ and $b$ . Direction is given by the right-hand rule.

Example:

The cross product of any two parallel vectors is the zero vector, since $sin 0 = 0$ .

📐 Worked Example

Find the magnitude of the cross product of $a = 2 i$ and $b = 3 j$ , and state its direction.

1
First find the angle between $a$ (x-axis) and $b$ (y-axis): $θ = 9 0^{\circ}$ , so $sin θ = 1$ .
2
$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = (2)(3)(1) = 6$
3
Apply the right-hand rule: index along x, middle along y, thumb points along the positive z-axis. Direction = positive z-axis.

2. Calculating Cross Product in Component Form★★★☆☆HL only⏱ 6 min

For vectors given in component form $a = a_{1} i + a_{2} j + a_{3} k$ and $b = b_{1} i + b_{2} j + b_{3} k$ , we use a 3x3 determinant to calculate the cross product:

\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\vec{i} - (a_1b_3 - a_3b_1)\vec{j} + (a_1b_2 - a_2b_1)\vec{k}

📐 Worked Example

Calculate $a \times b$ where $a = 123$ and $b = 456$ .

1
Substitute the components into the determinant formula:
2
$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}$
3
Expand the determinant along the first row:
4
$= \vec{i}(2 \cdot 6 - 3 \cdot 5) - \vec{j}(1 \cdot 6 - 3 \cdot 4) + \vec{k}(1 \cdot 5 - 2 \cdot 4)$
5
$= -3\vec{i} + 6\vec{j} - 3\vec{k}$
6
Check your result: the cross product must be perpendicular to both inputs. The dot product with $a$ is $(- 3) (1) + (6) (2) + (- 3) (3) = 0$ , which confirms correctness.

3. Key Algebraic Properties★★★☆☆HL only⏱ 4 min

Cross product follows these core properties that you will use to simplify expressions and solve problems:

Anti-commutativity: $a \times b = - b \times a$
Distributive over addition: $a \times (b + c) = a \times b + a \times c$
Scalar associativity: $(k a) \times b = k (a \times b)$
Self cross product: $a \times a = 0$
Unit vector products: $i \times j = k, j \times k = i, k \times i = j$

✓ Quick check

Test your understanding of cross product properties:

What is the value of $j \times i$ ?
- $k$
- $- k$
- $0$
- $1$
Reveal answer
$-\vec{k}$ —
Correct: cross product is anti-commutative, so $j \times i = - i \times j = - k$ .
If $a \times b = 0$ , what does this tell us about $a$ and $b$ ?
- They are perpendicular
- They are parallel
- They are unit vectors
- They have equal magnitude
Reveal answer
They are parallel —
Right! $∣ a \times b ∣ = 0$ implies $sin θ = 0$ , so $θ = 0$ or $π$ , meaning vectors are parallel.

4. Geometric Applications★★★★☆HL only⏱ 5 min

The magnitude of the cross product has a direct geometric meaning: it equals the area of the parallelogram spanned by the two input vectors. This leads to a range of practical applications in 3D geometry.

📘 Definition

Area Formulas

Area of parallelogram with adjacent sides $a, b$ = $∣ a \times b ∣$ . Area of triangle with adjacent sides $a, b$ = $\frac{1}{2} ∣ a \times b ∣$ . Cross product also gives a normal vector perpendicular to a plane containing the two vectors.

📐 Worked Example

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

1
Find two sides of the triangle starting from the same vertex A:
2
$\vec{AB} = (-1, 2, 0), \quad \vec{AC} = (-1, 0, 3)$
3
Calculate the cross product:
4
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{vmatrix} = 6\vec{i} + 3\vec{j} + 2\vec{k}$
5
Find the magnitude of the cross product:
6
$|\vec{AB} \times \vec{AC}| = \sqrt{6^2 + 3^2 + 2^2} = 7$
7
Area of the triangle is half the magnitude:
8
$\text{Area} = \frac{7}{2} = 3.5 \text{ square units}$

5. Common Pitfalls

Wrong move:

Forgetting the negative sign on the $j$ term when expanding the cross product determinant

Why:

The cofactor of the middle element in the first row of a determinant has a negative sign, which is often missed

Correct move:

Always use the pattern: $+ i (...) - j (...) + k (...)$ when expanding the cross product determinant

Wrong move:

Confusing cross product with dot product, giving a scalar result instead of a vector

Why:

Cross product produces a vector, while dot product produces a scalar, mixing them leads to lost marks

Correct move:

Always check the question: questions asking for perpendicular vectors or area use cross product, questions asking for angles use dot product

Wrong move:

Forgetting to halve the cross product magnitude when calculating triangle area

Why:

Cross product magnitude gives the area of the full parallelogram, not half for the triangle

Correct move:

Always confirm the shape you are calculating area for, remember $Area_{triangle} = \frac{1}{2} ∣ a \times b ∣$

Wrong move:

Claiming $a \times b = 0$ means one vector is the zero vector

Why:

Cross product is zero for any two parallel vectors, not just when one is zero

Correct move:

If $a \times b = 0$ , conclude that $a$ and $b$ are parallel (collinear if they share a common point)

Wrong move:

Using cross product to find a perpendicular vector for two 2D vectors

Why:

Cross product is only defined for two 3D vectors

Correct move:

For a 2D vector $(a, b)$ , a perpendicular vector is simply $(- b, a)$ , no cross product needed

6. Quick Reference Cheatsheet

Concept	Formula
Magnitude of cross product	$∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$
Component form calculation	$i a_{1} b_{1} j a_{2} b_{2} k a_{3} b_{3}$
Anti-commutativity	$a \times b = - b \times a$
Parallel vectors condition	$a \times b = 0$
Area of parallelogram	$∣ a \times b ∣$
Area of triangle	$\frac{1}{2} ∣ a \times b ∣$
Normal vector to a plane	$n = a \times b$ for two vectors in the plane

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find area of a triangle
2022 · 1
Check collinearity of points
2023 · 2
Find normal to a plane

Going deeper

official syllabusIB AA HL Topic 3 SyllabusOfficial curriculum reference

What's Next

Now that you have mastered the cross product of vectors, you can apply this core HL tool to more advanced 3D geometry problems involving lines and planes. Cross product is the foundation for finding equations of planes, calculating the angle between two planes, finding distances from points to planes, and calculating volumes of parallelepipeds using the scalar triple product, which combines dot product and cross product. Mastery of cross product calculations is essential for almost all 3D geometry questions that appear in Paper 1 and Paper 2 of IB AA HL, so it is important to practice expanding determinants and checking your results for perpendicularity.