MathsCBSE Class 12 · Vector Algebra

Vector (Cross) Product: Determinant Method and Areas (Class 12)

Updated 15 October 2026 · Fact-checked

The vector product a × b is a vector of magnitude |a||b| sin θ, perpendicular to both, by the right-hand rule. In components it is a 3 × 3 determinant with î, ĵ, k̂ in the first row. |a × b| is the area of the parallelogram on a and b; half of it is the triangle's area.

The parallelogram's area equals |a × b| = |a||b| sin θ, and a × b points out of the page; î → ĵ → k̂ is positive, reversed is negative.

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 from $a$ to $b$ ; the thumb points along $\overset{n}{^}$ ). Unlike the dot product, the result is a vector.

Unit vector products. $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ (cyclic order), the reverse order gives a minus sign, 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 $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ ,
$a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3} = (a_{2} b_{3} - a_{3} b_{2}) \hat{i} - (a_{1} b_{3} - a_{3} b_{1}) \hat{j} + (a_{1} b_{2} - a_{2} b_{1}) \hat{k}$ .
Do not forget the minus sign on the $\hat{j}$ term.

Properties.

Anti-commutative: $b \times a = - (a \times b)$ .
Distributive: $a \times (b + c) = a \times b + a \times c$ .
$a \times b = 0$ for non-zero vectors exactly when they are parallel (collinear); in particular $a \times a = 0$ .
$∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ (since $sin^{2} + cos^{2} = 1$ ).

Perpendicular unit vectors. The unit vectors perpendicular to both $a$ and $b$ are $\pm \frac{a \times b}{∣ a \times b ∣}$ . A vector of magnitude $k$ perpendicular to both is $\pm k$ times this.

Areas.

Parallelogram with adjacent sides $a$ , $b$ : area $= ∣ a \times b ∣$ .
Parallelogram with diagonals $d_{1}$ , $d_{2}$ : area $= \frac{1}{2} ∣ d_{1} \times d_{2} ∣$ .
Triangle ABC: area $= \frac{1}{2} ∣ A B \times A C ∣$ . Three points are collinear exactly when this is zero.

The cross product also appears in physics as torque, $τ = r \times F$ . In the syllabus, it is NCERT Section 10.6.3; the scalar triple product was dropped. Area questions with the cross product appear almost every year as 2-mark items.

Vector (Cross) Product of Vectors: formulas to remember

Definition: $a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}$; $\overset{n}{^}$ perpendicular to both, right-hand rule.
Determinant form: $a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3}$; Expand along the first row; middle term has 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.
Parallel vectors: $a \times b = 0 ⟺ a ∥ b$; For non-zero vectors.
Areas: $parallelogram = ∣ a \times b ∣, triangle = \frac{1}{2} ∣ A B \times A C ∣$; With diagonals: $\frac{1}{2} ∣ d_{1} \times d_{2} ∣$ .
Lagrange's identity: $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$; Links dot and cross products.

How to solve Vector (Cross) Product of Vectors questions

Use this routine for cross products, perpendicular vectors and area questions.

1Write both vectors in component form, in the order given.
2Set up the $3 \times 3$ determinant with $\hat{i}, \hat{j}, \hat{k}$ in the first row and expand, keeping the minus sign on $\hat{j}$ .
3Check: the result should have zero dot product with each original vector.
4For a unit perpendicular vector, divide by $∣ a \times b ∣$ and give both signs.
5For areas, choose the right formula (adjacent sides, diagonals, or a triangle from three points) and take the magnitude.
6State the answer in square units, simplifying surds.

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.

Torque as a cross product

Suppose a mechanic uses a spanner 0.3 m long (along the x-axis from the bolt) and pushes on its end with a 50 N force in the xy-plane.

Find the magnitude of the torque $r \times F$ when the force is perpendicular to the spanner, and when it makes an angle of $30°$ with it.

Try it first, then show the solutionSolution

$∣ r \times F ∣ = ∣ r ∣∣ F ∣ sin θ$ with $∣ r ∣ = 0.3$ m, $∣ F ∣ = 50$ N.
Perpendicular ( $θ = 90°$ ): $0.3 \times 50 \times 1 = 15$ N m.
At $30°$ : $0.3 \times 50 \times \frac{1}{2} = 7.5$ N m.

Answer: 15 N m when perpendicular; 7.5 N m at $30°$ .

Memory hook: Push at right angles to the spanner: the sine is 1 and the turning effect is greatest.

Area of a triangle in space

Imagine a triangular sail with corners at $A (0, 0, 0)$ , $B (4, 0, 0)$ and $C (1, 0, 3)$ , in metres.

Find the area of cloth needed for the sail.

Try it first, then show the solutionSolution

$A B = 4 \hat{i}$ , $A C = \hat{i} + 3 \hat{k}$ .
$A B \times A C = \hat{i} 41 \hat{j} 00 \hat{k} 03 = 0 \hat{i} - 12 \hat{j} + 0 \hat{k}$ .
Area $= \frac{1}{2} \times 12 = 6$ m² (check: base 4, height 3).

Answer: 6 m²

Memory hook: Triangle area = half the cross product of two sides from one corner.

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.

Cyclic order i, j, k
Going round $\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}$ gives a plus; going backwards gives a minus.
Use it when: Unit-vector products.
Minus in the middle
In the determinant expansion, the $\hat{j}$ term carries a minus sign.
Use it when: Avoiding the commonest sign error.
Dot to check
Your cross product must be perpendicular to both vectors: dot it with each and get 0.
Use it when: A 20-second check.
Sine for cross, cosine for dot
Cross product magnitude uses sine (area); dot product uses cosine (projection).
Use it when: Lagrange's identity and MCQs.

Vector (Cross) Product of Vectors: solved examples

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

Example 1: A cross product and its magnitude · 2 marks · Very short answerBoard level

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

Try it first, then show the solutionSolution

$a \times b = \hat{i} (2 - 3) - \hat{j} (- 4 - 1) + \hat{k} (6 + 1) = - \hat{i} + 5 \hat{j} + 7 \hat{k}$ . (1 mark)
$∣ a \times b ∣ = 1 + 25 + 49 = 75 = 53$ . (1 mark)

Answer: $- \hat{i} + 5 \hat{j} + 7 \hat{k}$ ; magnitude $53$

Example 2: Area of a triangle from its vertices · 2 marks · Very short answerBoard level

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

Try it first, then show the solutionSolution

$A B = 2 \hat{i} + \hat{j} - \hat{k}$ , $A C = \hat{i} + 2 \hat{j} + \hat{k}$ ; $A B \times A C = \hat{i} (1 + 2) - \hat{j} (2 + 1) + \hat{k} (4 - 1) = 3 \hat{i} - 3 \hat{j} + 3 \hat{k}$ . (1 mark)
Area $= \frac{1}{2} 9 + 9 + 9 = \frac{3 3}{2}$ square units. (1 mark)

Answer: $\frac{3 3}{2}$ square units

Example 3: Unit vectors perpendicular to two vectors · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

$a \times b = \hat{i} (1 + 6) - \hat{j} (3 + 4) + \hat{k} (9 - 2) = 7 \hat{i} - 7 \hat{j} + 7 \hat{k}$ . (1 mark)
$∣ a \times b ∣ = 73$ . (1 mark)
Unit vectors: $\pm \frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$ . Check: $(3, 1, - 2) \cdot (1, - 1, 1) = 0$ and $(2, 3, 1) \cdot (1, - 1, 1) = 0$ . (1 mark)

Answer: $\pm \frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$

Example 4: Using Lagrange's identity · 2 marks · Very short answerHarder board level

If $∣ a ∣ = 2$ , $∣ b ∣ = 5$ and $a \cdot b = 8$ , find $∣ a \times b ∣$ .

Try it first, then show the solutionSolution

$∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = 100 - 64 = 36$ . (1 mark)
$∣ a \times b ∣ = 6$ . (1 mark)

Answer: 6

Example 5: Three vectors adding to zero · 3 marks · Short answerHOTS / competency

If $a + b + c = 0$ , prove that $a \times b = b \times c = c \times a$ , and interpret the result for a triangle.

Try it first, then show the solutionSolution

Cross $a$ with both sides: $a \times a + a \times b + a \times c = 0$ , so $a \times b = - a \times c = c \times a$ . (1 mark)
Cross $b$ : $b \times a + b \times c = 0$ , so $b \times c = - b \times a = a \times b$ . (1 mark)
If $a, b, c$ are the sides of a triangle taken in order, each cross product has magnitude twice the triangle's area, so the area is the same whichever two sides you use. (1 mark)

Answer: $a \times b = b \times c = c \times a$ .

Example 6: Case study: a parallelogram-shaped plot · 4 marks · Case-basedHOTS / competency

Suppose a plot ABCD is a parallelogram with $A B = 4 \hat{i} + 2 \hat{j}$ and $A D = \hat{i} + 3 \hat{j}$ (in tens of metres).
(i) Find the area of the plot. (1 mark)
(ii) Find the diagonals $A C$ and $B D$ . (1 mark)
(iii) Verify the area using the diagonals. (1 mark)
(iv) Find the area of triangle ABC. (1 mark)

Try it first, then show the solutionSolution

(i) $A B \times A D = (4 \cdot 3 - 2 \cdot 1) \hat{k} = 10 \hat{k}$ ; area $= 10$ square units (1000 m²). (1 mark)
(ii) $A C = A B + A D = 5 \hat{i} + 5 \hat{j}$ ; $B D = A D - A B = - 3 \hat{i} + \hat{j}$ . (1 mark)
(iii) $\frac{1}{2} ∣ A C \times B D ∣ = \frac{1}{2} ∣5 (1) - 5 (- 3) ∣ = \frac{1}{2} (20) = 10$ . (1 mark)
(iv) Triangle ABC is half the parallelogram: 5 square units (500 m²). (1 mark)

Answer: (i) 10 sq. units (ii) $5 \hat{i} + 5 \hat{j}$ , $- 3 \hat{i} + \hat{j}$ (iii) 10 (iv) 5 sq. units.

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

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

Mistake: Missing the minus sign on the ĵ term
The determinant expansion alternates signs.
Wrong working
$a \times b = (2 - 3) \hat{i} + (- 4 - 1) \hat{j} + (6 + 1) \hat{k}$ .
Right working
$= (2 - 3) \hat{i} - (- 4 - 1) \hat{j} + (6 + 1) \hat{k} = - \hat{i} + 5 \hat{j} + 7 \hat{k}$ .
Do this: Write ' $- \hat{j} (\dots)$ ' explicitly.
Usually costs: The answer and any area based on it.
Mistake: Forgetting ½ for a triangle
The cross product gives the parallelogram's area.
Wrong working
Area of triangle $= ∣ A B \times A C ∣ = 33$ .
Right working
$= \frac{1}{2} ∣ A B \times A C ∣ = \frac{3 3}{2}$ .
Do this: Write the area formula before substituting.
Usually costs: 1 mark.
Mistake: Giving only one unit perpendicular vector
There are two, opposite in direction.
Wrong working
$\frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$ .
Right working
$\pm \frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$ .
Do this: Write $\pm$ unless a direction is specified.
Usually costs: Half a mark.
Mistake: Swapping the order of vectors
$b \times a = - a \times b$ .
Wrong working
$a \times b$ computed with $b$ in row 2.
Right working
Put $a$ in row 2 and $b$ in row 3.
Do this: Keep the order of the question.
Usually costs: The sign of the answer.

Exam tips

Area of a parallelogram or triangle using the cross product appeared in the 2023 to 2026 papers as 2-mark questions; show the determinant.
Check your cross product by dotting it with both vectors.
For magnitude-only data, Lagrange's identity is quicker than finding angles.
Zero cross product means the vectors are parallel; use this to find unknowns.
The scalar triple product $[a b c]$ is not in the 2026-27 syllabus.

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
The value of $(\hat{j} \times \hat{k}) \cdot \hat{i} + \hat{j} \cdot \hat{j}$ is:
(a) 0
(b) 1
(c) $- 1$
(d) 2
Hint
$\hat{j} \times \hat{k} = \hat{i}$ and $\hat{j} \cdot \hat{j} = 1$ .
Answer
(d) $\hat{i} \cdot \hat{i} + 1 = 2$ .
Q2 · 1 mark · Very short answerBoard level
If $∣ a ∣ = 3$ , $∣ b ∣ = 2$ and the angle between them is $\frac{π}{6}$ , find $∣ a \times b ∣$ .
Hint
$∣ a ∣∣ b ∣ sin θ$ .
Answer
$3 \times 2 \times \frac{1}{2} = 3$ .
Q3 · 2 marks · Very short answerBoard level
Find the area of the parallelogram whose diagonals are $2 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} - \hat{j} + 2 \hat{k}$ .
Hint
Area $= \frac{1}{2} ∣ d_{1} \times d_{2} ∣$ .
Answer
$d_{1} \times d_{2} = 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$ , magnitude $33$ ; area $= \frac{3 3}{2}$ square units.
Q4 · 2 marks · Very short answerHarder board level
Find $λ$ if $(2 \hat{i} + 6 \hat{j} + 14 \hat{k}) \times (\hat{i} - λ \hat{j} + 7 \hat{k}) = 0$ .
Hint
Zero cross product means parallel: components proportional.
Answer
$\frac{2}{1} = \frac{6}{- λ} = \frac{14}{7}$ , so $- λ = 3$ and $λ = - 3$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the angle between non-zero vectors $a$ and $b$ is $\frac{π}{4}$ , then $∣ a \times b ∣ = a \cdot b$ .
Reason (R): $sin \frac{π}{4} = cos \frac{π}{4}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare $∣ a ∣∣ b ∣ sin θ$ with $∣ a ∣∣ b ∣ cos θ$ .
Answer
(a). Both equal $∣ a ∣∣ b ∣ \cdot \frac{1}{2}$ , because $sin \frac{π}{4} = cos \frac{π}{4}$ .
Q6 · 3 marks · Short answerHarder board level
Find a vector of magnitude 6 perpendicular to both $2 \hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - \hat{k}$ .
Hint
The cross product has magnitude $32$ .
Answer
Cross product $= \hat{i} (1 - 1) - \hat{j} (- 2 - 1) + \hat{k} (2 + 1) = 3 \hat{j} + 3 \hat{k}$ , magnitude $32$ . Required vectors: $\pm \frac{6}{3 2} (3 \hat{j} + 3 \hat{k}) = \pm (32 \hat{j} + 32 \hat{k})$ .
Q7 · 2 marks · Very short answerHOTS / competency
If $∣ a ∣ = 5$ , $∣ b ∣ = 13$ and $∣ a \times b ∣ = 25$ , find $a \cdot b$ .
Hint
Use Lagrange's identity.
Answer
$(a \cdot b)^{2} = 25 \times 169 - 625 = 3600$ , so $a \cdot b = \pm 60$ .

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 $\hat{j} \times \hat{i}$ ?Show answer
$- \hat{k}$ .
2.Magnitude of $a \times b$ ?Show answer
$∣ a ∣∣ b ∣ sin θ$ .
3.Area of triangle ABC with vectors?Show answer
$\frac{1}{2} ∣ A B \times A C ∣$ .
4.When is $a \times b = 0$ for non-zero vectors?Show answer
When they are parallel.
5.Area of a parallelogram from diagonals?Show answer
$\frac{1}{2} ∣ d_{1} \times d_{2} ∣$ .
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

What is the difference between dot and cross products?

The dot product gives a number, |a||b| cos θ, used for angles and projections. The cross product gives a vector of magnitude |a||b| sin θ perpendicular to both, used for areas and perpendicular directions.

How do I find a unit vector perpendicular to two vectors?

Compute a × b, find its magnitude, and divide. Both the result and its negative are unit vectors perpendicular to a and b.

Is the cross product commutative?

No. b × a = −(a × b). The magnitudes are equal but the directions are opposite.

Which NCERT section covers the vector product?

Section 10.6.3 (Vector (or cross) product of two vectors) of Chapter 10 in NCERT Mathematics Part II for Class 12.

How can I check that three points are collinear using the cross product?

Compute AB × AC. If it is the zero vector, the area of triangle ABC is zero and the points are collinear.