PhysicsISC Class 11 · Motion in a Plane

Dot and Cross Products: Angle, Work, Torque and Area

Updated 17 October 2026 · Fact-checked

The scalar (dot) product A·B = AB cosθ = AₓBₓ + A_yB_y + A_zB_z is a scalar and gives work when A is force and B is displacement. The vector (cross) product A×B has magnitude AB sinθ and a direction given by the right-hand rule; it gives torque and the area of a parallelogram.

Parallelepiped on vectors a, b, c with arrows for a×b, b×c and the triple products a·b×c and a×b·c, in English labels. — Notice that a×b points perpendicular to the plane of a and b, and dotting it with c gives the parallelepiped's volume.
Picture: Brews ohare · CC BY-SA 3.0 · Wikimedia Commons

Dot product uses the part of B along A (B cosθ); cross product uses the part at right angles (B sinθ), which gives the parallelogram area.

Illustration drawn for QuestionBuddy.

Understand Dot and Cross Products of Vectors

Dot product. A·B = AB cosθ. It is commutative. It is zero for perpendicular vectors, equal to AB for parallel vectors and equal to A² when B = A. With unit vectors: i·i = j·j = k·k = 1 and i·j = j·k = k·i = 0. In components, A·B = AₓBₓ + A_yB_y + A_zB_z. The angle follows from cosθ = A·B/(AB). Work done by a constant force is W = F·s, and power is P = F·v.

Cross product. A×B = AB sinθ n̂, where n̂ is perpendicular to both vectors, given by the right-hand screw rule. It is not commutative: B×A = −(A×B). It is zero for parallel vectors. For unit vectors: i×j = k, j×k = i, k×i = j, and the reverse order gives a minus sign; i×i = 0. In component form, A×B is the determinant of the 3 × 3 array with rows (i j k), (Aₓ A_y A_z) and (Bₓ B_y B_z).

Applications. The magnitude |A×B| is the area of the parallelogram on A and B; half of it is the area of the triangle. Torque is τ = r×F, and angular momentum is L = r×p. A unit vector perpendicular to both A and B is (A×B)/|A×B|.

Checks. (A·B)² + |A×B|² = A²B² always holds. If A·B = 0 and A×B = 0 for non-zero vectors, there is an error.

Dot and Cross Products of Vectors: formulas and key facts

Dot product: $A \cdot B = A B cos θ ∣ A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$; A scalar; zero if perpendicular.
Cross product: $∣ A \times B ∣ = A B s in θ ∣ A \times B = (A_{y} B_{z} - A_{z} B_{y}) i + (A_{z} B_{x} - A_{x} B_{z}) j + (A_{x} B_{y} - A_{y} B_{x}) k$; Direction by the right-hand rule; zero if parallel.
Angle between vectors: $cos θ = A \cdot B / (A B)$; From the dot product.
Identity: $(A \cdot B)^{2} + ∣ A \times B ∣^{2} = A^{2} B^{2}$; A useful check.

Dot and Cross Products 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 on a spanner

Suppose a mechanic applies a force F = 5 N along the z-axis (k direction) at the end of a spanner whose end is at the position r = (3i + 4j) m from the nut.

Find the torque about the nut.

Try it first, then show the solutionSolution

τ = r × F = (3i + 4j) × 5k = 15(i × k) + 20(j × k).
i × k = −j and j × k = i, so τ = −15j + 20i = (20i − 15j) N m.
|τ| = √(400 + 225) = 25 N m.

Answer: τ = (20i − 15j) N m, magnitude 25 N m.

Memory hook: Learn the cycle i→j→k→i is positive, the reverse is negative.

Tricks to remember Dot and Cross Products of Vectors

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

Cycle for the cross product
i → j → k → i gives plus; reversing the order gives minus.
Use it when: Getting the signs of i×k and k×j right.
Dot kills perpendicular, cross kills parallel
Perpendicular: A·B = 0. Parallel: A×B = 0.
Use it when: Testing a given pair of vectors.

Dot and Cross Products of Vectors: solved examples

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

Example 1: Perpendicular vectors · 1 mark · MCQBoard level

If A·B = 0 for two non-zero vectors, then the vectors are:
(a) parallel
(b) perpendicular
(c) antiparallel
(d) equal

Try it first, then show the solutionSolution

A·B = AB cosθ = 0 with A and B non-zero requires cosθ = 0, so θ = 90°.

Answer: (b) perpendicular

Example 2: Dot and cross products · 3 marks · Short answerBoard level

For A = 2i + 3j + k and B = i − 2j + 4k, find A·B and A×B and the magnitude of A×B.

Try it first, then show the solutionSolution

A·B = 2 − 6 + 4 = 0, so the vectors are perpendicular. (1 mark)
A×B = i(3 × 4 − 1 × (−2)) − j(2 × 4 − 1 × 1) + k(2 × (−2) − 3 × 1) = 14i − 7j − 7k. (1 mark)
|A×B| = √(196 + 49 + 49) = √294 = 17.1. Check: AB = √14 × √21 = √294. (1 mark)

Answer: A·B = 0; A×B = 14i − 7j − 7k; magnitude 17.1.

Example 3: Angle between vectors · 3 marks · Short answerHarder board level

Find the angle between A = 2i + j + k and B = i + 2j − k.

Try it first, then show the solutionSolution

A·B = 2 + 2 − 1 = 3. (1 mark)
A = √6 and B = √6, so AB = 6. (1 mark)
cosθ = 3/6 = 0.5, so θ = 60°. (1 mark)

Answer: 60°

Example 4: Work by a force · 3 marks · Short answerHarder board level

A constant force F = (3i + 2j − k) N displaces a body by s = (2i + j + 4k) m. Find the work done and the angle between F and s.

Try it first, then show the solutionSolution

W = F·s = 6 + 2 − 4 = 4 J. (1 mark)
F = √14, s = √21, so Fs = √294 = 17.15. (1 mark)
cosθ = 4/17.15 = 0.233, so θ = 76.5°. (1 mark)

Answer: W = 4 J; θ ≈ 76.5°.

Example 5: Perpendicular unit vector and area · 5 marks · Long answerHOTS / competency

For A = i + 2j + 3k and B = 3i + 2j + k, find a unit vector perpendicular to both, and the area of the triangle with A and B as two sides.

Try it first, then show the solutionSolution

A×B = i(2 × 1 − 3 × 2) − j(1 × 1 − 3 × 3) + k(1 × 2 − 2 × 3) = −4i + 8j − 4k. (2 marks)
|A×B| = √(16 + 64 + 16) = √96 = 9.80. (1 mark)
Unit vector = (−4i + 8j − 4k)/√96 = (−i + 2j − k)/√6. Check: dot with A = −1 + 4 − 3 = 0 and with B = −3 + 4 − 1 = 0. (1 mark)
Triangle area = ½ × 9.80 = 4.90 square units. (1 mark)

Answer: Unit vector (−i + 2j − k)/√6 (or its negative); area 4.90 square units.

Where students lose marks in Dot and Cross Products of Vectors

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

Mistake: Wrong sign in the j term of the determinant
Forgetting the minus sign in front of the j component.
Do this: Expand the determinant as +i(...) − j(...) + k(...).
Usually costs: 1 to 2 marks.
Mistake: Using B×A for A×B
The result has the opposite sign.
Do this: Keep the order of the vectors as in the question; B×A = −A×B.
Usually costs: 1 mark.
Mistake: Calling work a vector
W = F·s is a scalar.
Do this: Write the dot product, and give the answer in joules without direction.
Usually costs: 1 mark.

Exam tips

Write both vectors in i, j, k form first and list the components.
Check a cross product by taking its dot product with each original vector; it must be zero.
Quote the formula (dot or cross) before substituting.
Give units: J for work and N m for torque.

Practice questions on Dot and Cross Products 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 i·(j×k) is:
(a) 0
(b) 1
(c) −1
(d) i
Hint
j×k = i.
Answer
(b) 1
Q2 · 2 marks · Short answerBoard level
Find A·B for A = 3i − 2j + k and B = 2i + 3j + k.
Hint
Multiply matching components and add.
Answer
6 − 6 + 1 = 1.
Q3 · 3 marks · Short answerHarder board level
Two vectors have magnitudes 3 and 4 with an angle of 60° between them. Verify that (A·B)² + |A×B|² = A²B².
Hint
Find A·B and |A×B| separately.
Answer
A·B = 12 cos60° = 6, so (A·B)² = 36. |A×B| = 12 sin60° = 10.39, so its square = 108. Sum = 144 = 3² × 4².
Q4 · 3 marks · Short answerHarder board level
A force F = 4k N acts at the point r = (2i + 3j) m. Find the torque about the origin and its magnitude.
Hint
τ = r × F; i×k = −j, j×k = i.
Answer
τ = 8(i×k) + 12(j×k) = −8j + 12i = (12i − 8j) N m; magnitude √(144 + 64) = 14.4 N m.
Q5 · 3 marks · Short answerHarder board level
Find λ if A = 2i + λj + k and B = 4i − 2j − 2k are perpendicular. Can they ever be parallel?
Hint
Set A·B = 0; for parallel, compare ratios of components.
Answer
A·B = 8 − 2λ − 2 = 0 → λ = 3. For parallel, 2/4 = λ/(−2) = 1/(−2) requires 1/2 = −1/2, which is false, so they are never parallel.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a force F = (10i + 5j) N moves a trolley through s = (3i + 4j) m.
(i) Find the work done. (1 mark)
(ii) Find the component of the force along s. (1 mark)
(iii) Find the angle between F and s. (1 mark)
(iv) If F acts at the point r = s from the origin, find the torque about the origin. (1 mark)
Hint
The component of F along s is F·s/|s|.
Answer
(i) 30 + 20 = 50 J. (ii) |s| = 5 m, so 50/5 = 10 N. (iii) |F| = √125 = 11.18; cosθ = 50/(11.18 × 5) = 0.894 → θ = 26.6°. (iv) τ_z = 3 × 5 − 4 × 10 = −25, so τ = −25k N m.

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.A·B in terms of angle?Show answer
AB cosθ.
2.|A×B| in terms of angle?Show answer
AB sinθ.
3.What is i×j?Show answer
k.
4.Which product gives the area of a parallelogram?Show answer
The cross product, |A×B|.

Dot and Cross Products of Vectors: frequently asked questions

Is the cross product commutative?

No. A×B = −(B×A), so the order matters. The dot product is commutative.

Why is work a scalar?

Work depends on the component of force along the displacement, found by the dot product, which is a scalar with no direction.

What does the right-hand rule say?

Curl the fingers of the right hand from the first vector towards the second through the smaller angle; the thumb points along the direction of the cross product.