MathsISC Class 12 · Vector Algebra

Vector (Cross) Product: Determinant Method, Areas and Parallel Vectors

Updated 15 October 2026 · Fact-checked

The vector (cross) product a × b is a vector of magnitude |a||b|sin θ, perpendicular to both a and b by the right-hand rule. It is computed as a determinant. Its magnitude is the area of the parallelogram on a and b, and it is zero exactly when a and b are parallel.

a × b stands perpendicular to the plane of a and b, and its length |a||b| sin θ equals the parallelogram's area.

Illustration drawn for QuestionBuddy.

Understand Vector (Cross) Product of Vectors

Definition. For vectors $a$ and $b$ with angle $θ$ between them,
$a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^},$
where $\overset{n}{^}$ is the unit vector perpendicular to both $a$ and $b$ , chosen by the right-hand rule: curl the fingers of the right hand from $a$ towards $b$ , and the thumb points along $\overset{n}{^}$ . Unlike the dot product, the result is a vector.

Unit vectors. $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ (in cyclic order), while reversing the order gives the negative: $\hat{j} \times \hat{i} = - \hat{k}$ . And $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$ .

Determinant form. For $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ :
$a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3} = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k} .$
Watch the minus sign on the $\hat{j}$ term; it is where most slips happen.

Properties.

Anti-commutative: $b \times a = - (a \times b)$ .
Distributive: $a \times (b + c) = a \times b + a \times c$ .
$a \times a = 0$ .
Parallel vectors: for non-zero vectors, $a \times b = 0$ if and only if $a ∥ b$ (since $sin θ = 0$ ). So if $a \times b = a \times c$ , then $a \times (b - c) = 0$ : either $b = c$ or $a$ is parallel to $b - c$ (the ISC 2024 2-mark question; the examiners noted candidates who did not use the distributive law to reach this).
Lagrange's identity: $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ , because $sin^{2} θ + cos^{2} θ = 1$ .

Geometrical significance.

Area of the parallelogram with adjacent sides $a$ , $b$ : $∣ a \times b ∣$ .
Area of the triangle with two sides $a$ , $b$ : $\frac{1}{2} ∣ a \times b ∣$ . For vertices $A, B, C$ : $\frac{1}{2} ∣ A B \times A C ∣$ .
Area of a parallelogram from its diagonals $d_{1}$ , $d_{2}$ : $\frac{1}{2} ∣ d_{1} \times d_{2} ∣$ .
Unit vector perpendicular to both $a$ and $b$ : $\pm \frac{a \times b}{∣ a \times b ∣}$ (two answers, opposite directions).
Collinear points: $A$ , $B$ , $C$ are collinear if $A B \times A C = 0$ .

Physics link. Torque is $τ = r \times F$ , and the force on a moving charge in a magnetic field is $q v \times B$ , so the cross product reappears throughout Class 12 Physics. (Proving general geometrical theorems with vectors is excluded from the 2027 syllabus; the cross product is used here for calculation.)

Vector (Cross) Product of Vectors: formulas to remember

Definition: $a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}$; n̂ perpendicular to both, by the right-hand rule.
Determinant form: $\hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3}$; Expand along the first row; the ĵ term takes a minus sign.
Unit vectors: $\hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{k} = \hat{i}, \hat{k} \times \hat{i} = \hat{j}$; Reverse order gives the negative; î × î = 0.
Anti-commutative: $b \times a = - a \times b$; Reversing the order reverses the direction.
Parallel vectors: $a \times b = 0 ⟺ a ∥ b$; For non-zero vectors.
Areas: $parallelogram = ∣ a \times b ∣, triangle = \frac{1}{2} ∣ a \times b ∣$; From diagonals: ½|d₁ × d₂|.
Unit vector perpendicular to a and b: $\pm \frac{a \times b}{∣ a \times b ∣}$; Two opposite answers.
Lagrange's identity: $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$; From sin²θ + cos²θ = 1.

How to solve Vector (Cross) Product of Vectors questions

Most cross-product questions ask for the product itself, a perpendicular unit vector, an area or a parallel/collinear test.

1Write both vectors in component form; for points, form side vectors as (terminal − initial).
2Set up the $3 \times 3$ determinant with $\hat{i}, \hat{j}, \hat{k}$ in the top row.
3Expand along the top row, remembering the minus sign on the $\hat{j}$ term; write each $2 \times 2$ minor before simplifying.
4Check your answer: its dot product with each original vector must be 0.
5For an area, find the magnitude; halve it for a triangle (or for a parallelogram given by diagonals).
6For a perpendicular unit vector, divide by the magnitude and give both signs.
7For parallel or collinear tests, show the cross product is the zero vector.

Vector (Cross) Product of Vectors in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area of a triangle by the cross product

Suppose a triangular sail is fixed at three points with coordinates $(0, 0, 0)$ , $(3, 0, 0)$ and $(0, 2, 4)$ , in metres.

Find the area of the sail.

Try it first, then show the solutionSolution

Side vectors from the origin: $a = 3 \hat{i}$ , $b = 2 \hat{j} + 4 \hat{k}$ .
$a \times b = (0 \cdot 4 - 0 \cdot 2) \hat{i} - (3 \cdot 4 - 0) \hat{j} + (3 \cdot 2 - 0) \hat{k} = - 12 \hat{j} + 6 \hat{k}$ .
$∣ a \times b ∣ = 144 + 36 = 180 = 65$ .
Area $= \frac{1}{2} \times 65 = 35 \approx 6.71$ m².

Answer: $35 \approx 6.71$ m².

Memory hook: Half the length of the cross product of two sides gives the triangle area.

Torque as a cross product

Imagine pushing a door at its handle, 0.8 m from the hinge along $\hat{i}$ , with a force of 30 N straight into the door along $\hat{j}$ .

Find the torque $τ = r \times F$ about the hinge.

Try it first, then show the solutionSolution

$r = 0.8 \hat{i}$ , $F = 30 \hat{j}$ .
$τ = 0.8 \times 30 (\hat{i} \times \hat{j}) = 24 \hat{k}$ N m.
The torque points along the hinge axis, which is exactly the axis the door turns about.

Answer: $24 \hat{k}$ N m (24 N m about the hinge).

Memory hook: Turning effect = position × force, pointing along the axis of turning.

Tricks to remember Vector (Cross) Product of Vectors

Short memory aids. Say each one aloud once, then use it on a question the same day.

i-j-k clockwise
Write i, j, k round a circle: going forwards gives +, backwards gives −.
Use it when: Cross products of unit vectors.
Minus in the middle
In the determinant expansion, the ĵ term always carries a minus sign.
Use it when: Avoiding the commonest sign error.
Zero cross means parallel
Dot zero: perpendicular. Cross zero: parallel.
Use it when: Parallel and collinear tests.
Check by dotting
a × b must be perpendicular to a and to b: both dot products must be 0.
Use it when: Catching arithmetic slips.

Vector (Cross) Product of Vectors: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Unit vector products · 1 mark · MCQBoard level

The value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{i}) \cdot \hat{k}$ is:
(a) 2
(b) 1
(c) 0
(d) $- 2$

Try it first, then show the solutionSolution

$\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = - \hat{k}$ , so the sum is $1 - 1 = 0$ .

Answer: (c) 0.

Example 2: A cross product by determinant · 2 marks · Short answerBoard level

Find $a \times b$ and $∣ a \times b ∣$ for $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + \hat{k}$ .

Try it first, then show the solutionSolution

$\hat{i} 12 \hat{j} 2 - 1 \hat{k} 31 = (2 + 3) \hat{i} - (1 - 6) \hat{j} + (- 1 - 4) \hat{k} = 5 \hat{i} + 5 \hat{j} - 5 \hat{k}$ . (1½ marks)
$∣ a \times b ∣ = 53$ . (½ mark)

Answer: $5 \hat{i} + 5 \hat{j} - 5 \hat{k}$ ; magnitude $53$ .

Example 3: A unit vector perpendicular to two vectors · 3 marks · Short answerHarder board level

Find a unit vector perpendicular to both $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$ .

Try it first, then show the solutionSolution

$a \times b = (3 - 2) \hat{i} - (3 - 1) \hat{j} + (2 - 1) \hat{k} = \hat{i} - 2 \hat{j} + \hat{k}$ . (1½ marks)
Check: $(\hat{i} - 2 \hat{j} + \hat{k}) \cdot a = 0$ and $\cdot b = 1 - 4 + 3 = 0$ . (½ mark)
Magnitude $6$ , so the unit vectors are $\pm \frac{1}{6} (\hat{i} - 2 \hat{j} + \hat{k})$ . (1 mark)

Answer: $\pm \frac{1}{6} (\hat{i} - 2 \hat{j} + \hat{k})$ .

Example 4: Area of a triangle from vertices · 3 marks · Short answerHarder board level

Find the area of the triangle with vertices $A (1, 1, 2)$ , $B (2, 3, 5)$ and $C (1, 5, 5)$ .

Try it first, then show the solutionSolution

$A B = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $A C = 4 \hat{j} + 3 \hat{k}$ . (1 mark)
$A B \times A C = (6 - 12) \hat{i} - (3 - 0) \hat{j} + (4 - 0) \hat{k} = - 6 \hat{i} - 3 \hat{j} + 4 \hat{k}$ . (1 mark)
Area $= \frac{1}{2} 36 + 9 + 16 = \frac{61}{2}$ sq units. (1 mark)

Answer: $\frac{61}{2}$ sq units.

Example 5: Making two vectors parallel · 2 marks · Short answerHarder board level

Find $λ$ and $μ$ if $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + λ \hat{j} + μ \hat{k}) = 0$ .

Try it first, then show the solutionSolution

A zero cross product means the vectors are parallel, so their components are proportional: $\frac{1}{2} = \frac{λ}{6} = \frac{μ}{27}$ . (1 mark)
$λ = 3$ , $μ = \frac{27}{2}$ . (1 mark)

Answer: $λ = 3$ , $μ = \frac{27}{2}$ .

Example 6: Diagonals, collinearity and Lagrange's identity · 5 marks · Long answerHOTS / competency

(i) Find the area of the parallelogram whose diagonals are $\hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} - 2 \hat{j} + \hat{k}$ . (2 marks)
(ii) Show that $A (2, - 1, 3)$ , $B (3, 1, 2)$ and $C (5, 5, 0)$ are collinear. (1 mark)
(iii) If $∣ a ∣ = 2$ , $∣ b ∣ = 5$ and $∣ a \times b ∣ = 8$ , find $a \cdot b$ . (2 marks)

Try it first, then show the solutionSolution

(i) $d_{1} \times d_{2} = (2 + 6) \hat{i} - (1 - 9) \hat{j} + (- 2 - 6) \hat{k} = 8 \hat{i} + 8 \hat{j} - 8 \hat{k}$ , of magnitude $83$ . Area $= \frac{1}{2} \times 83 = 43$ .
(ii) $A B = \hat{i} + 2 \hat{j} - \hat{k}$ and $A C = 3 \hat{i} + 6 \hat{j} - 3 \hat{k} = 3 A B$ , so $A B \times A C = 0$ : collinear.
(iii) Lagrange: $(a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - ∣ a \times b ∣^{2} = 100 - 64 = 36$ .
So $a \cdot b = \pm 6$ (the angle may be acute or obtuse).

Answer: (i) $43$ sq units; (ii) collinear; (iii) $a \cdot b = \pm 6$ .

Where students lose marks in Vector (Cross) Product of Vectors

The slips examiners see most often, and what to write instead.

Mistake: Sign error on the ĵ term
The middle term of the determinant expansion carries a minus sign.
Wrong working
$(a_{1} b_{3} - a_{3} b_{1}) \hat{j}$ .
Right working
$- (a_{1} b_{3} - a_{3} b_{1}) \hat{j}$ .
Do this: Write the full expansion pattern before substituting numbers.
Usually costs: 1 mark, and every later part that uses the vector.
Mistake: Not using the distributive law
In ISC 2024 candidates failed to rewrite a × b = a × c as a × (b − c) = 0.
Do this: Move everything to one side and factor out the common vector.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the ½ for a triangle
The cross product magnitude gives the parallelogram area.
Do this: Triangle = ½|AB × AC|; parallelogram from diagonals = ½|d₁ × d₂| too.
Usually costs: 1 mark.
Mistake: Giving only one unit vector
There are two unit vectors perpendicular to a plane, in opposite directions.
Do this: Write ± before the answer.
Usually costs: ½ to 1 mark.

Exam tips

Show the determinant set-up; it earns the method mark even if a minor goes wrong.
Check your cross product by dotting it with one of the original vectors.
Keep magnitudes as surds ( $61$ ) unless asked for decimals.
For collinearity, a zero cross product (or one vector a scalar multiple of the other) is enough; state the conclusion.
Remember the 2027 syllabus excludes proofs of geometrical theorems by vectors; expect calculation and short property proofs instead.

Practice questions on Vector (Cross) Product of Vectors

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For any vector $a$ , $a \times a$ equals:
(a) $∣ a ∣^{2}$
(b) 1
(c) $0$
(d) $\overset{a}{^}$
Hint
The angle is zero.
Answer
(c) $0$ .
Q2 · 2 marks · Short answerBoard level
Find the area of the parallelogram with adjacent sides $\hat{i} + \hat{j} - \hat{k}$ and $2 \hat{i} - \hat{j} + \hat{k}$ .
Hint
Area $= ∣ a \times b ∣$ .
Answer
$a \times b = (1 - 1) \hat{i} - (1 + 2) \hat{j} + (- 1 - 2) \hat{k} = - 3 \hat{j} - 3 \hat{k}$ ; area $32$ sq units.
Q3 · 3 marks · Short answerHarder board level
If $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , find a unit vector perpendicular to both $a + b$ and $a - b$ .
Hint
$a + b = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ , $a - b = - \hat{j} - 2 \hat{k}$ .
Answer
Cross product $= - 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ , magnitude $26$ . Unit vectors $\pm \frac{1}{6} (- \hat{i} + 2 \hat{j} - \hat{k})$ .
Q4 · 2 marks · Short answerHarder board level
If $∣ a ∣ = 2$ , $∣ b ∣ = 7$ and $a \cdot b = 7$ , find $∣ a \times b ∣$ .
Hint
Lagrange's identity.
Answer
$∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = 196 - 49 = 147$ , so $∣ a \times b ∣ = 73$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $a \times b = - b \times a$ .
Reason (R): The vector product is commutative.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Reversing the order reverses the right-hand rule.
Answer
(c). A is true; the vector product is anti-commutative, so R is false.
Q6 · 3 marks · Short answerHOTS / competency
Using vectors, show that $A (1, - 2, - 8)$ , $B (5, 0, - 2)$ and $C (11, 3, 7)$ are collinear, and find the ratio in which $B$ divides $A C$ .
Hint
Compare $A B$ and $B C$ .
Answer
$A B = 4 \hat{i} + 2 \hat{j} + 6 \hat{k}$ , $B C = 6 \hat{i} + 3 \hat{j} + 9 \hat{k} = \frac{3}{2} A B$ , so $A B \times B C = 0$ : collinear. $A B : B C = 2 : 3$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a mechanic applies a force $F = 20 \hat{j} + 10 \hat{k}$ N at the end of a spanner, at position $r = 0.5 \hat{i}$ m from the bolt.
(i) Find the torque $τ = r \times F$ . (2 marks)
(ii) Find its magnitude. (1 mark)
(iii) Which component of a force applied at the same point would produce no torque at all? (1 mark)
Hint
Use $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{i} \times \hat{k} = - \hat{j}$ .
Answer
(i) $0.5 \times 20 (\hat{i} \times \hat{j}) + 0.5 \times 10 (\hat{i} \times \hat{k}) = 10 \hat{k} - 5 \hat{j}$ N m. (ii) $125 = 55 \approx 11.2$ N m. (iii) A component along $\hat{i}$ (along the spanner), since $\hat{i} \times \hat{i} = 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the magnitude of $a \times b$ ?Show answer
$∣ a ∣∣ b ∣ sin θ$ .
2.Find $\hat{k} \times \hat{j}$ .Show answer
$- \hat{i}$ .
3.When is $a \times b = 0$ for non-zero vectors?Show answer
When they are parallel.
4.Area of a triangle with sides $a$ and $b$ ?Show answer
$\frac{1}{2} ∣ a \times b ∣$ .
5.Is the cross product commutative?Show answer
No: $b \times a = - a \times b$ .
6.State Lagrange's identity.Show answer
$∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ .

Vector (Cross) Product of Vectors: frequently asked questions

How is the cross product different from the dot product?

The dot product gives a number, |a||b|cos θ; the cross product gives a vector of size |a||b|sin θ perpendicular to both. The dot product vanishes for perpendicular vectors, the cross product for parallel ones.

How do I find the direction of a × b?

Use the right-hand rule: curl the fingers from a towards b through the smaller angle; the thumb points along a × b. In component form, the determinant gives the direction automatically.

Why are there two unit vectors perpendicular to two given vectors?

Both a × b and b × a are perpendicular to a and b, and they point in opposite directions. So the answer is ±(a × b)/|a × b|.

How do I find the area of a triangle with given vertices?

Form two side vectors from one vertex, such as AB and AC, find their cross product, and take half its magnitude.

Is the vector product in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus includes the cross product, its geometrical significance and properties, including areas of triangles and parallelograms and collinear vectors, while excluding proofs of geometrical theorems by vectors.