MathsCBSE Class 12 · Vector Algebra

Scalar (Dot) Product of Vectors: Angle, Projection and Perpendicularity (Class 12)

Updated 15 October 2026 · Fact-checked

The scalar product of 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 gives the angle (cos θ = a · b/(|a||b|)), tests perpendicularity (a · b = 0) and gives the projection of a on b as a · b/|b|.

Drop a perpendicular from the tip of a onto b: the green length is a·b/|b|, and when θ is 90° the projection and a·b both become zero.

Illustration drawn for QuestionBuddy.

Understand Scalar (Dot) Product of Vectors

Definition. For non-zero 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 number (scalar), not a vector. If either vector is zero, the product is 0.

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

Properties.

Commutative: $a \cdot b = b \cdot a$ ; distributive: $a \cdot (b + c) = a \cdot b + a \cdot c$ .
$a \cdot a = ∣ a ∣^{2}$ , so $∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b$ .
$(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ .
Perpendicular non-zero vectors: $a \cdot b = 0$ . The converse: if $a \cdot b = 0$ , then $a = 0$ , $b = 0$ or $a ⊥ b$ .
Sign of the dot product: positive for an acute angle, zero for a right angle, negative for an obtuse angle.

Angle between two vectors. $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ ; leave the answer as $θ = cos^{- 1} (\dots)$ when it is not a standard angle.

Projection. The (scalar) projection of $a$ on $b$ is $\frac{a \cdot b}{∣ b ∣} = a \cdot \hat{b}$ , the length of the shadow of $a$ on the line of $b$ (negative if the angle is obtuse). The projection vector is $(a \cdot \hat{b}) \hat{b}$ .

Using magnitudes only. Many questions give magnitudes and a relation such as $a + b + c = 0$ . Square both sides using $v \cdot v = ∣ v ∣^{2}$ : for example $∣ c ∣^{2} = ∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ , which gives $a \cdot b$ and then the angle.

In physics. Work done by a constant force $F$ through a displacement $d$ is $W = F \cdot d$ .

This is NCERT Section 10.6.1 to 10.6.2 and is one of the most asked vector topics (angle, projection and perpendicularity questions appear every year).

Scalar (Dot) Product of Vectors: formulas to remember

Definition: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$; A scalar; $0 \leq θ \leq π$ .
Component form: $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$; Multiply matching components and add.
Angle: $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$; Leave non-standard answers as $cos^{- 1} (\dots)$ .
Perpendicular: $a ⊥ b ⟺ a \cdot b = 0$; For non-zero vectors.
Projection of a on b: $\frac{a \cdot b}{∣ b ∣}$; Divide by the magnitude of the vector you project onto.
Magnitude identities: $∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b$; Also $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ .
Work done: $W = F \cdot d$; Force times displacement in the direction of the force.

How to solve Scalar (Dot) Product of Vectors questions

Use this routine for angle, projection, perpendicularity and magnitude questions.

1Write each vector in component form and compute $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ .
2For an angle, find $∣ a ∣$ and $∣ b ∣$ and use $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ .
3For perpendicularity with an unknown, set the dot product to 0 and solve.
4For a projection of $a$ on $b$ , divide $a \cdot b$ by $∣ b ∣$ (not $∣ a ∣$ ).
5For magnitude-only questions, square the given relation using $v \cdot v = ∣ v ∣^{2}$ and solve for the unknown dot product.
6State the final answer with units (for work, joules) or as an angle.

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 you pull a suitcase 20 m along a level airport floor with a force of 50 N through its handle, which makes an angle of $60°$ with the floor.

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

Try it first, then show the solutionSolution

Take the floor as the x-axis: $F = 50 cos 60° \hat{i} + 50 sin 60° \hat{j} = 25 \hat{i} + 253 \hat{j}$ N; $d = 20 \hat{i}$ m.
$W = F \cdot d = 25 \times 20 + 253 \times 0 = 500$ J.
Only the component of the force along the floor does work; the upward part does none.

Answer: 500 J

Memory hook: Work counts only the part of the force that points the way you move.

Dot product as a weighted total

Imagine a shopping list with quantities $q = (2, 3, 1)$ for pens, notebooks and a file, and prices $p = (40, 25, 100)$ rupees.

Interpret $q \cdot p$ and compute it.

Try it first, then show the solutionSolution

$q \cdot p = 2 (40) + 3 (25) + 1 (100)$ .
$= 80 + 75 + 100 = 255$ .
Each term is (quantity × price), so the sum is the total bill.

Answer: Rs 255, the total cost.

Memory hook: Multiply matching parts and add: that is a dot product, and a shopping bill.

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.

Dot gives a number
The dot product of two vectors is a plain number; if your answer has $\hat{i}$ in it, you have done something else.
Use it when: Catching cross-product confusion.
Zero dot means right angle
Perpendicular? Set the dot product to zero.
Use it when: Finding unknowns.
Project onto: divide by its length
Projection of a ON b: divide by |b|, the vector you land on.
Use it when: Projection questions.
Square to bring in the dot
Given only magnitudes? Square the vector equation; the cross terms become dot products.
Use it when: Questions like $a + b + c = 0$ .

Scalar (Dot) Product of Vectors: solved examples

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

Example 1: Angle between two vectors · 2 marks · Very short answerBoard level

Find the angle between $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 2 \hat{k}$ .

Try it first, then show the solutionSolution

$a \cdot b = 2 - 2 + 4 = 4$ ; $∣ a ∣ = 3$ , $∣ b ∣ = 3$ . (1 mark)
$cos θ = \frac{4}{9}$ , so $θ = cos^{- 1} \frac{4}{9}$ . (1 mark)

Answer: $θ = cos^{- 1} \frac{4}{9}$

Example 2: Projection · 2 marks · Very short answerBoard level

Find the projection of $a = 3 \hat{i} - \hat{j} + 2 \hat{k}$ on $b = 2 \hat{i} + 2 \hat{j} + \hat{k}$ .

Try it first, then show the solutionSolution

$a \cdot b = 6 - 2 + 2 = 6$ ; $∣ b ∣ = 4 + 4 + 1 = 3$ . (1 mark)
Projection $= \frac{6}{3} = 2$ . (1 mark)

Answer: 2

Example 3: Perpendicular vectors with an unknown · 3 marks · Short answerHarder board level

Find $λ$ so that $a = λ \hat{i} + 2 \hat{j} + \hat{k}$ and $b = 4 \hat{i} - 9 \hat{j} + 2 \hat{k}$ are perpendicular. For this $λ$ , find $∣ a + b ∣$ .

Try it first, then show the solutionSolution

$a \cdot b = 4 λ - 18 + 2 = 0 \Rightarrow λ = 4$ . (1 mark)
$∣ a ∣^{2} = 16 + 4 + 1 = 21$ , $∣ b ∣^{2} = 16 + 81 + 4 = 101$ . (1 mark)
$∣ a + b ∣^{2} = 21 + 101 + 2 (0) = 122$ , so $∣ a + b ∣ = 122$ . (1 mark)

Answer: $λ = 4$ ; $∣ a + b ∣ = 122$

Example 4: Magnitudes of combinations · 2 marks · Very short answerHarder board level

If $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $a ⊥ b$ , find $∣ a + b ∣$ and $∣2 a - b ∣$ .

Try it first, then show the solutionSolution

$∣ a + b ∣^{2} = 9 + 16 + 0 = 25 \Rightarrow ∣ a + b ∣ = 5$ . (1 mark)
$∣2 a - b ∣^{2} = 4 (9) + 16 - 4 (0) = 52 \Rightarrow ∣2 a - b ∣ = 213$ . (1 mark)

Answer: 5 and $213$

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

If $a + b + c = 0$ with $∣ a ∣ = 2$ , $∣ b ∣ = 3$ and $∣ c ∣ = 19$ , find the angle between $a$ and $b$ , and the value of $a \cdot b + b \cdot c + c \cdot a$ .

Try it first, then show the solutionSolution

$c = - (a + b)$ , so $∣ c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ . (1 mark)
$19 = 4 + 9 + 2 a \cdot b \Rightarrow a \cdot b = 3$ . (1 mark)
$cos θ = \frac{3}{2 \times 3} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$ . (1 mark)
Square $a + b + c = 0$ : $∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a) = 0$ . (1 mark)
$a \cdot b + b \cdot c + c \cdot a = - \frac{4 + 9 + 19}{2} = - 16$ . (1 mark)

Answer: $θ = \frac{π}{3}$ ; the sum is $- 16$ .

Example 6: Case study: pushing a crate · 4 marks · Case-basedHOTS / competency

Suppose a worker applies a constant force $F = 3 \hat{i} + 2 \hat{j} - \hat{k}$ (newtons) to move a crate from $A (1, 1, 2)$ to $B (4, 3, 5)$ (metres).
(i) Find the displacement vector. (1 mark)
(ii) Find the work done. (1 mark)
(iii) Find the component of the force along the displacement. (1 mark)
(iv) Is the angle between the force and the displacement acute or obtuse? Why? (1 mark)

Try it first, then show the solutionSolution

(i) $d = A B = 3 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . (1 mark)
(ii) $W = F \cdot d = 9 + 4 - 3 = 10$ J. (1 mark)
(iii) $\frac{F \cdot d}{∣ d ∣} = \frac{10}{22}$ N $\approx 2.13$ N. (1 mark)
(iv) $F \cdot d = 10 > 0$ , so $cos θ > 0$ and the angle is acute ( $cos θ = \frac{10}{14 22}$ ). (1 mark)

Answer: (i) $3 \hat{i} + 2 \hat{j} + 3 \hat{k}$ (ii) 10 J (iii) $\frac{10}{22}$ N (iv) acute.

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

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

Mistake: Dividing by the wrong magnitude for a projection
The projection of $a$ on $b$ uses $∣ b ∣$ .
Wrong working
Projection of $a$ on $b$ $= \frac{a \cdot b}{∣ a ∣}$ .
Right working
$= \frac{a \cdot b}{∣ b ∣}$ .
Do this: Divide by the length of the vector you project onto.
Usually costs: The answer.
Mistake: Writing a·a = a² as a vector
$a \cdot a$ is the number $∣ a ∣^{2}$ .
Wrong working
$∣ a + b ∣^{2} = a^{2} + b^{2}$ .
Right working
$∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b$ .
Do this: Expand with the dot product and keep the cross term.
Usually costs: 1 to 2 marks.
Mistake: Arithmetic slips in components
Negative components are easy to mishandle.
Wrong working
$(2, - 1, 2) \cdot (1, 2, 2) = 2 + 2 + 4 = 8$ .
Right working
$= 2 - 2 + 4 = 4$ .
Do this: Write each product with its sign before adding.
Usually costs: The final answer.
Mistake: Giving a decimal angle without cos⁻¹
Non-standard angles are expected as $cos^{- 1} (\dots)$ .
Wrong working
$θ = 0.4$ .
Right working
$θ = cos^{- 1} \frac{4}{9}$ .
Do this: Leave exact forms.
Usually costs: Half a mark.

Exam tips

Dot-product questions (angle, projection, perpendicularity) appeared in every board paper from 2023 to 2026; write the formula before substituting.
For 'find λ for perpendicular vectors', one line $a \cdot b = 0$ and one line of solving is enough.
When only magnitudes are given, square the given vector relation.
Projection can be negative; do not drop the sign.
In case studies with forces, work done is $F \cdot d$ with $d$ = final minus initial position.

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) $\hat{i}$
Hint
Perpendicular unit vectors have dot product 0.
Answer
(b) $1 + 0 = 1$ .
Q2 · 2 marks · Very short answerBoard level
Find the angle between $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ .
Hint
Dot product 1, magnitudes $2$ .
Answer
$cos θ = \frac{1}{2 2} = \frac{1}{2}$ , so $θ = \frac{π}{3}$ .
Q3 · 2 marks · Very short answerBoard level
If $∣ a ∣ = 2$ , $∣ b ∣ = 3$ and $a \cdot b = 3$ , find the angle between them and $∣ a - b ∣$ .
Hint
$∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$ .
Answer
$cos θ = \frac{3}{6} = \frac{1}{2}$ , $θ = \frac{π}{3}$ . $∣ a - b ∣^{2} = 4 + 9 - 6 = 7$ , so $∣ a - b ∣ = 7$ .
Q4 · 3 marks · Short answerHarder board level
Prove that if $∣ a ∣ = ∣ b ∣$ , then $a + b$ is perpendicular to $a - b$ . Interpret this for a rhombus.
Hint
Expand $(a + b) \cdot (a - b)$ .
Answer
$(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2} = 0$ , so they are perpendicular (when non-zero). In a rhombus with sides $a$ and $b$ , the diagonals $a + b$ and $a - b$ are perpendicular.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $a \cdot b = 0$ , then $a = 0$ or $b = 0$ .
Reason (R): $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ .
(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
What if $cos θ = 0$ ?
Answer
(d). $\hat{i} \cdot \hat{j} = 0$ with both vectors non-zero, so A is false; R is the definition and is true.
Q6 · 2 marks · Very short answerHarder board level
If $a$ is a unit vector and $(x - a) \cdot (x + a) = 15$ , find $∣ x ∣$ .
Hint
The left side is $∣ x ∣^{2} - ∣ a ∣^{2}$ .
Answer
$∣ x ∣^{2} - 1 = 15 \Rightarrow ∣ x ∣ = 4$ .
Q7 · 3 marks · Short answerHOTS / competency
Using vectors, prove that the angle in a semicircle is a right angle.
Hint
Take the centre O as origin, A and B with position vectors $a$ and $- a$ , and P on the circle with $∣ p ∣ = ∣ a ∣$ .
Answer
$P A = a - p$ and $P B = - a - p$ . Then $P A \cdot P B = - (a - p) \cdot (a + p) = - (∣ a ∣^{2} - ∣ p ∣^{2}) = 0$ . So $∠ A P B = 90°$ .

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.Component formula for $a \cdot b$ ?Show answer
$a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ .
2.Condition for perpendicular non-zero vectors?Show answer
$a \cdot b = 0$ .
3.Projection of $a$ on $b$ ?Show answer
$\frac{a \cdot b}{∣ b ∣}$ .
4.What is $a \cdot a$ ?Show answer
$∣ a ∣^{2}$ .
5.Sign of $a \cdot b$ for an obtuse angle?Show answer
Negative.
6.Expand $∣ a - b ∣^{2}$ .Show answer
$∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b$ .

Scalar (Dot) Product of Vectors: frequently asked questions

Is the dot product a vector or a scalar?

A scalar. a · b = |a||b| cos θ is a single number, which is why it is also called the scalar product.

How do I find the angle between two vectors?

Compute a · b in components, find |a| and |b|, and use cos θ = a · b/(|a||b|). Write θ = cos⁻¹ of that value if it is not a standard angle.

What is the difference between projection and projection vector?

The projection of a on b is the number a · b/|b|. The projection vector is that number times the unit vector of b, so it points along b.

Which NCERT section covers the dot product?

Section 10.6 (Product of Two Vectors), with 10.6.1 (Scalar or dot product) and 10.6.2 (Projection of a vector on a line), in NCERT Mathematics Part II for Class 12.

If a · b = a · c, can I cancel a?

No. It only means a · (b − c) = 0, so b − c may be perpendicular to a without being zero.