MathsISC Class 12 · Vector Algebra

Scalar (Dot) Product of Vectors: Angle, Perpendicularity and Identities

Updated 15 October 2026 · Fact-checked

The scalar (dot) product of vectors a and b is a·b = |a||b|cos θ, where θ is the angle between them; in components it is a₁b₁ + a₂b₂ + a₃b₃. It is a number, not a vector. Non-zero vectors are perpendicular exactly when a·b = 0, and cos θ = a·b/(|a||b|) gives the angle.

Understand Scalar (Dot) Product of Vectors

Definition. For vectors $a$ and $b$ with angle $θ$ between them ( $0 \leq θ \leq π$ ),
$a \cdot b = ∣ a ∣∣ b ∣ cos θ .$
The result is a scalar. It is positive when the angle is acute, zero when the vectors are perpendicular, and negative when the angle is obtuse.

Component form. Since $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$ , for $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ :
$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} .$

Angle between two vectors.
$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣} = \frac{a _{1} b _{1} + a _{2} b _{2} + a _{3} b _{3}}{a _{1}^{2} + a _{2}^{2} + a _{3}^{2} b _{1}^{2} + b _{2}^{2} + b _{3}^{2}} .$

Perpendicular vectors. Non-zero vectors are perpendicular if and only if $a \cdot b = 0$ . This turns geometric conditions into equations: to make $2 \hat{i} + λ \hat{j} + \hat{k}$ perpendicular to $\hat{i} - 2 \hat{j} + 3 \hat{k}$ , set $2 - 2 λ + 3 = 0$ .

Properties.

Commutative: $a \cdot b = b \cdot a$ .
Distributive: $a \cdot (b + c) = a \cdot b + a \cdot c$ .
Scalars move out: $(λ a) \cdot b = λ (a \cdot b)$ .
$a \cdot a = ∣ a ∣^{2}$ , written $a^{2}$ or $a^{2}$ .

Key identities. These come straight from the properties and are used constantly:

$∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$
$∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$
$(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$

Adding the first two gives the parallelogram law, $∣ a + b ∣^{2} + ∣ a - b ∣^{2} = 2 (∣ a ∣^{2} + ∣ b ∣^{2})$ , which was the assertion in an ISC 2025 item. (Using vectors to prove general geometrical theorems is excluded from the 2027 syllabus; these identities are used for calculation.)

When three vectors add to zero. If $a + b + c = 0$ , then $c = - (a + b)$ , and squaring gives $∣ c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ . This finds angles between the sides of a triangle from their lengths.

Physical meaning. Work done by a constant force $F$ moving an object through displacement $d$ is $W = F \cdot d$ : only the component of the force along the motion does work. The related idea of projection is developed on the projection topic page.

Scalar (Dot) Product of Vectors: formulas to remember

Definition: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$; A scalar.
Component form: $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$; Multiply matching components and add.
Angle: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$; 0 ≤ θ ≤ π.
Perpendicular: $a ⊥ b ⟺ a \cdot b = 0$; For non-zero vectors.
Unit vectors: $\hat{i} \cdot \hat{i} = 1, \hat{i} \cdot \hat{j} = 0$; And similarly for the other pairs.
Square of a sum: $∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b$; Use to find magnitudes of sums and differences.
Difference of squares: $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$; Zero when |a| = |b|.
Work done: $W = F \cdot d$; F is the force, d the displacement.

How to solve Scalar (Dot) Product of Vectors questions

Most dot-product questions are one of four kinds: compute, find an angle, impose perpendicularity, or use an identity. Use these steps.

1Write each vector in $\hat{i}, \hat{j}, \hat{k}$ form; if points are given, form the vectors as (terminal − initial).
2Compute $a \cdot b$ by multiplying matching components and adding.
3For an angle, find both magnitudes and use $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ ; give $θ$ as $cos^{- 1} (\dots)$ if it is not standard.
4For perpendicularity, set the dot product equal to zero and solve for the unknown.
5If only magnitudes and a relation are given, square the relation using $∣ u ∣^{2} = u \cdot u$ and expand with the distributive law.
6For a triangle given by vertices, a zero dot product of two side vectors shows a right angle.
7Check signs: an obtuse angle must come from a negative dot product.

Scalar (Dot) 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.

Work done as a dot product

Suppose a child pulls a sledge 20 m along level snow with a rope that makes an angle of $3 0^{\circ}$ with the ground, keeping a tension of 50 N in the rope.

Write the force and displacement as vectors and find the work done.

Try it first, then show the solutionSolution

Take $\hat{i}$ along the ground and $\hat{j}$ vertically: $F = 50 cos 3 0^{\circ} \hat{i} + 50 sin 3 0^{\circ} \hat{j} = 253 \hat{i} + 25 \hat{j}$ .
$d = 20 \hat{i}$ .
$W = F \cdot d = 253 \times 20 + 25 \times 0 = 5003 \approx 866$ J.

Answer: $5003 \approx 866$ joules; the upward part of the pull does no work.

Memory hook: Only the part of the force along the motion does work: that is the dot product.

Angle between two vectors

Imagine a delivery drone flying with velocity $3 \hat{i} + 4 \hat{j}$ m/s (east and north) while the wind blows with velocity $2 \hat{i} - \hat{j}$ m/s.

Find the angle between the drone's velocity and the wind.

Try it first, then show the solutionSolution

$v \cdot w = 6 - 4 = 2$ ; $∣ v ∣ = 5$ , $∣ w ∣ = 5$ .
$cos θ = \frac{2}{5 5} \approx 0.179$ .
$θ = cos^{- 1} \frac{2}{5 5} \approx 79. 7^{\circ}$ .

Answer: About $79. 7^{\circ}$ : the wind is almost a crosswind.

Memory hook: cos θ = dot product over the product of lengths.

Tricks to remember Scalar (Dot) Product of Vectors

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

Same with same is one
î·î = ĵ·ĵ = k̂·k̂ = 1; different unit vectors give 0.
Use it when: Expanding dot products of component vectors.
Zero means right angle
A dot product of 0 between non-zero vectors means they are perpendicular.
Use it when: Perpendicularity and right-triangle questions.
Square it to open it
Given magnitudes of sums or differences? Square them: |u|² = u·u.
Use it when: Problems with |a + b| or a + b + c = 0.
Sign tells the angle
Positive: acute. Zero: right. Negative: obtuse.
Use it when: Quick checks on answers.

Scalar (Dot) Product of Vectors: solved examples

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

Example 1: A direct dot product · 1 mark · MCQBoard level

If $a = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $b = \hat{i} + 4 \hat{j} - \hat{k}$ , then $a \cdot b$ is:
(a) 5
(b) $- 5$
(c) 9
(d) $- 1$

Try it first, then show the solutionSolution

$2 (1) + (- 1) (4) + 3 (- 1) = 2 - 4 - 3 = - 5$ .

Answer: (b) $- 5$ .

Example 2: Angle between two vectors · 2 marks · Short answerBoard level

Find the angle between $a = \hat{i} + \hat{j}$ and $b = \hat{j} + \hat{k}$ .

Try it first, then show the solutionSolution

$a \cdot b = 0 + 1 + 0 = 1$ ; $∣ a ∣ = ∣ b ∣ = 2$ . (1 mark)
$cos θ = \frac{1}{2}$ , so $θ = \frac{π}{3}$ . (1 mark)

Answer: $\frac{π}{3}$ ( $6 0^{\circ}$ ).

Example 3: Finding an unknown for perpendicularity · 2 marks · Short answerHarder board level

Find $λ$ so that $3 \hat{i} + λ \hat{j} - 2 \hat{k}$ and $\hat{i} + 4 \hat{j} + λ \hat{k}$ are perpendicular.

Try it first, then show the solutionSolution

Perpendicular means the dot product is zero: $3 + 4 λ - 2 λ = 0$ . (1 mark)
$2 λ = - 3$ , so $λ = - \frac{3}{2}$ . (1 mark)

Answer: $λ = - \frac{3}{2}$ .

Example 4: Using magnitudes only · 3 marks · Short answerHarder board level

If $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $∣ a + b ∣ = 5$ , find $a \cdot b$ , the angle between $a$ and $b$ , and $∣ a - b ∣$ .

Try it first, then show the solutionSolution

$∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ : $25 = 9 + 16 + 2 a \cdot b$ , so $a \cdot b = 0$ . (1 mark)
So the angle is $\frac{π}{2}$ . (1 mark)
$∣ a - b ∣^{2} = 9 + 16 - 0 = 25$ , so $∣ a - b ∣ = 5$ . (1 mark)

Answer: $a \cdot b = 0$ ; angle $9 0^{\circ}$ ; $∣ a - b ∣ = 5$ .

Example 5: Magnitudes of a sum and a difference · 3 marks · Short answerHarder board level

If $∣ a ∣ = 2$ , $∣ b ∣ = 3$ and the angle between $a$ and $b$ is $6 0^{\circ}$ , find $∣ a + b ∣$ and $∣ a - b ∣$ .

Try it first, then show the solutionSolution

$a \cdot b = 2 \times 3 \times cos 6 0^{\circ} = 3$ . (1 mark)
$∣ a + b ∣^{2} = 4 + 9 + 2 (3) = 19$ , so $∣ a + b ∣ = 19$ . (1 mark)
$∣ a - b ∣^{2} = 4 + 9 - 2 (3) = 7$ , so $∣ a - b ∣ = 7$ . (1 mark)

Answer: $∣ a + b ∣ = 19$ , $∣ a - b ∣ = 7$ .

Example 6: Three vectors adding to zero · 5 marks · Long answerHOTS / competency

Vectors $a$ , $b$ , $c$ satisfy $a + b + c = 0$ , with $∣ a ∣ = 3$ , $∣ b ∣ = 5$ and $∣ c ∣ = 7$ .
(i) Find the angle between $a$ and $b$ . (3 marks)
(ii) Find $a \cdot b + b \cdot c + c \cdot a$ . (2 marks)

Try it first, then show the solutionSolution

(i) $c = - (a + b)$ , so $∣ c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ : $49 = 9 + 25 + 2 a \cdot b$ , so $a \cdot b = 7.5$ .
$cos θ = \frac{7.5}{3 \times 5} = \frac{1}{2}$ , so $θ = 6 0^{\circ}$ .
(ii) Square $a + b + c = 0$ : $∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a) = 0$ .
$83 + 2 S = 0$ , so $S = - 41.5$ .

Answer: (i) $6 0^{\circ}$ ; (ii) $- \frac{83}{2}$ .

Where students lose marks in Scalar (Dot) Product of Vectors

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

Mistake: Writing the dot product as a vector
a·b is a number; writing 2î − 4ĵ − 3k̂ for it confuses it with the component-wise product.
Wrong working
$a \cdot b = 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$ .
Right working
$a \cdot b = 2 - 4 - 3 = - 5$ .
Do this: Add the products of matching components to get one number.
Usually costs: 1 mark.
Mistake: Calculation errors with signs
The ISC 2024 examiners noted calculation errors in dot- and cross-product questions.
Do this: Write each product with its sign before adding, for example (−1)(4) = −4.
Usually costs: 1 mark.
Mistake: Forming side vectors in the wrong order
AB = OB − OA, not OA − OB; the 2025 examiners noted this error.
Do this: Always write "terminal minus initial".
Usually costs: A sign error that can change an angle from acute to obtuse.
Mistake: Squaring magnitudes wrongly
|a + b| is not |a| + |b|; it must be squared with the dot product.
Do this: Use |a + b|² = |a|² + |b|² + 2a·b.
Usually costs: The whole question.

Exam tips

Show the dot product and both magnitudes on separate lines before writing $cos θ$ .
If the angle is not standard, leave it as $cos^{- 1} (\dots)$ ; do not convert to decimals unless asked.
For "show that the triangle is right-angled", one zero dot product is enough; say which vertex.
In identity proofs, cite commutativity when you combine $a \cdot b$ and $b \cdot a$ .
Remember $a^{2}$ means $a \cdot a = ∣ a ∣^{2}$ , a number.

Practice questions on Scalar (Dot) 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 $\hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{k}$ is:
(a) 0
(b) 1
(c) 2
(d) $- 1$
Hint
Same unit vectors give 1; different give 0.
Answer
(b) 1.
Q2 · 2 marks · Short answerBoard level
Find the angle between $\hat{i} - \hat{j}$ and $- \hat{i} + \hat{k}$ .
Hint
The dot product is negative.
Answer
Dot product $- 1$ ; magnitudes $2$ each; $cos θ = - \frac{1}{2}$ , so $θ = \frac{2 π}{3}$ ( $12 0^{\circ}$ ).
Q3 · 2 marks · Short answerHarder board level
If $∣ b ∣ = 2$ and $(a + b) \cdot (a - b) = 12$ , find $∣ a ∣$ .
Hint
Use the difference-of-squares identity.
Answer
$∣ a ∣^{2} - 4 = 12$ , so $∣ a ∣ = 4$ .
Q4 · 3 marks · Short answerHarder board level
Using vectors, show that the points $A (2, - 1, 1)$ , $B (1, - 3, - 5)$ and $C (3, - 4, - 4)$ form a right-angled triangle, and find the vertex with the right angle.
Hint
Compute the side vectors and their dot products.
Answer
$C A = - \hat{i} + 3 \hat{j} + 5 \hat{k}$ , $C B = - 2 \hat{i} + \hat{j} - \hat{k}$ . $C A \cdot C B = 2 + 3 - 5 = 0$ : right angle at $C$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $a$ and $b$ are non-zero vectors with $a \cdot b = 0$ , then $a ⊥ b$ .
Reason (R): $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ .
(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
What is θ when cos θ = 0?
Answer
(a). With non-zero magnitudes, $a \cdot b = 0$ gives $cos θ = 0$ , so $θ = 9 0^{\circ}$ .
Q6 · 3 marks · Short answerHOTS / competency
If $\overset{a}{^}$ and $\hat{b}$ are unit vectors inclined at angle $θ$ , prove that $∣ \overset{a}{^} - \hat{b} ∣ = 2 sin \frac{θ}{2}$ .
Hint
Square the magnitude.
Answer
$∣ \overset{a}{^} - \hat{b} ∣^{2} = 1 + 1 - 2 cos θ = 2 (1 - cos θ) = 4 sin^{2} \frac{θ}{2}$ . Since $0 \leq θ \leq π$ , $sin \frac{θ}{2} \geq 0$ , so $∣ \overset{a}{^} - \hat{b} ∣ = 2 sin \frac{θ}{2}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a crate is dragged by a constant force $F = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ newtons from the point $A (1, 0, 2)$ to $B (4, 2, 1)$ (in metres).
(i) Find the displacement vector. (1 mark)
(ii) Find the work done. (1 mark)
(iii) Find the angle between the force and the displacement. (1 mark)
(iv) Find the component of the force along the displacement. (1 mark)
Hint
Work $= F \cdot d$ .
Answer
(i) $d = 3 \hat{i} + 2 \hat{j} - \hat{k}$ . (ii) $9 + 8 - 5 = 12$ J. (iii) $cos θ = \frac{12}{5 2 \cdot 14} = \frac{6}{5 7} \approx 0.454$ , so $θ \approx 63. 0^{\circ}$ . (iv) $\frac{12}{14} \approx 3.21$ N.

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.Define $a \cdot b$ .Show answer
$∣ a ∣∣ b ∣ cos θ$ , where $θ$ is the angle between them.
2.Find $(2 \hat{i} + \hat{j}) \cdot (\hat{i} - 3 \hat{j})$ .Show answer
$2 - 3 = - 1$ .
3.When are two non-zero vectors perpendicular?Show answer
When their dot product is 0.
4.Expand $∣ a - b ∣^{2}$ .Show answer
$∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$ .
5.What is $\hat{j} \cdot \hat{j}$ ?Show answer
1.
6.Is the dot product commutative?Show answer
Yes: $a \cdot b = b \cdot a$ .

Scalar (Dot) Product of Vectors: frequently asked questions

What is the difference between the dot product and the cross product?

The dot product a·b = |a||b|cos θ is a scalar and measures how much two vectors point the same way. The cross product a × b is a vector of magnitude |a||b|sin θ, perpendicular to both. The dot product is zero for perpendicular vectors; the cross product is zero for parallel ones.

Can the dot product be negative?

Yes. It is negative when the angle between the vectors is obtuse, because cos θ is negative for angles between 90° and 180°.

How do I find the angle between two vectors?

Compute the dot product and both magnitudes, then use cos θ = a·b/(|a||b|) and take the inverse cosine. The angle always lies between 0 and π.

What does a·a equal?

It equals |a|², the square of the magnitude. This is why squaring expressions like |a + b| is done with dot products.

Is the scalar product in the ISC 2027 syllabus?

Yes. Unit 4, Vector Algebra, in the CISCE 2027 syllabus includes the scalar (dot) product, its geometrical meaning and properties, and the unit carries 5 marks in the paper.