NDA Maths · Vectors

Dot Product and Angle

The scalar that captures how aligned two vectors are — used to compute angles, test perpendicularity, and evaluate work done by a force.

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Why this matters

The dot product turns two vectors into a single number — and that number contains everything you'd ever want to know about how the two vectors RELATE. Are they perpendicular? At what angle do they meet? How aligned are they with each other? The five concepts below build that toolkit: starting with the formula itself (and its physical meaning as work done by a force, W = F · d), through the perpendicularity test and the angle formula, ending with the most-tested setup — "given a constraint on two vectors a and b, find the angle between them." 32 PYQs across 2017–2026 — the second-biggest Vectors subtopic; almost every paper has one — with a difficulty mix of 12 EASY + 16 MODERATE + 4 HARD.

Test yourself — a quick 12-question Vectors recall quiz, instant score.

Concept 1 of 5: Dot product — components form and work done

The dot product of two vectors is a single number that measures how much they overlap. Compute it by multiplying corresponding components and summing — no angles needed. Physically, when a constant force

\vec{F}

displaces a particle by

\vec{d}

, the work done is exactly

\vec{F}\cdot\vec{d}

Definition

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ , then $\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3$ . It is commutative, distributive over addition, and a scalar (not a vector). Work done by a constant force is $W = \vec{F}\cdot\vec{d}$ where $\vec{d}$ is the displacement.

Dot product (components form)

\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad W = \vec{F}\cdot\vec{d}

$a_i, b_i$ components of $\vec{a}, \vec{b}$ along $\hat{i}, \hat{j}, \hat{k}$
$W$ work done by a constant force $\vec{F}$ through displacement $\vec{d}$

Diagram · work = F · d = |F||d| cos θ

Only the part of the force along the displacement does work: W = F · d = |F||d| cos θ. A force perpendicular to the motion (θ = 90°) does zero work; one opposing it (θ > 90°) does negative work.

Worked example

A force

\vec{F} = 2\hat{i} + \hat{j} - \hat{k}

displaces a particle from

A(1, 0, 2)

B(3, 2, 1)

. Find the work done.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2017 · Q47Easy

Example 1 · Vectors · Dot Product and Angle

A force

\vec{F} = \hat{i}+3\hat{j}+2\hat{k}

acts on a particle to displace it from the point

A(\hat{i}+2\hat{j}-3\hat{k})

to the point

B(3\hat{i}-\hat{j}+5\hat{k})

. The work done by the force will be

Dot product gives a scalar; cross product gives a vector

An MCQ option that returns a vector for

\vec{a}\cdot\vec{b}

, or a scalar for

\vec{a}\times\vec{b}

, can be eliminated on type grounds alone. This dimension-check rejects ~25% of trap options.

Work done is signed — negative work is fine

If the force has any component opposite to the displacement, the dot product (and the work) can come out negative. Don't reach for absolute value automatically; the sign tells you whether the force is helping or hindering.

Drill 5 more on dot product — components form and work done

Concept 2 of 5: Perpendicularity Test

Two non-zero vectors are perpendicular precisely when their dot product is zero. This single test has three equivalent disguises that PYQs love to switch between:

\vec{a}\cdot\vec{b}=0

directly;

|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|

(the diagonals of a parallelogram are equal iff it's a rectangle); and

(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2

(the Pythagoras identity).

Definition

For non-zero $\vec{a}, \vec{b}$ , all three are equivalent — each says $\vec{a} \perp \vec{b}$ :

$\vec{a}\cdot\vec{b} = 0$
$|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$
$(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$

Equivalent perpendicularity statements

\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|

$\vec{a}\cdot\vec{b}$ scalar dot product
$|\vec{a}\pm\vec{b}|$ magnitudes of the diagonals of the parallelogram on $\vec{a}, \vec{b}$

Worked example

Find the value of

\lambda

so that

\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}

and

\vec{b} = \hat{i} + \lambda\hat{j} - 5\hat{k}

are perpendicular.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q70Easy

Example 2 · Vectors · Dot Product and Angle

What is the value of

\lambda

for which the vectors

3\hat{i}+4\hat{j}-\hat{k}

and

-2\hat{i}+\lambda\hat{j}+10\hat{k}

are perpendicular?

$|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$ means $\vec{a}\perp\vec{b}$ , not $\vec{a}=\vec{b}$

Square both sides:

|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2

collapses to

4\vec{a}\cdot\vec{b} = 0

. Geometrically: the two diagonals of a parallelogram have equal length iff the parallelogram is a rectangle.

Zero dot product needs both vectors non-zero

Technically

\vec{0}\cdot\vec{a} = 0

for any

\vec{a}

, but

\vec{0}

has no direction, so we don't call it perpendicular. PYQs assume non-zero vectors implicitly; double-check the hypothesis if a question opens with \"if non-zero...\".

Drill 5 more on perpendicularity test

Concept 3 of 5: Angle between two vectors via the dot-product formula

The dot product is the magnitude of one vector times the magnitude of the other times the cosine of the angle between them. Rearrange and the angle drops out: divide the dot product by the product of magnitudes. Sign of the result tells you acute (

>0

) versus obtuse (

<0

);

=0

is perpendicular, the special case above.

Definition

For non-zero $\vec{a}, \vec{b}$ at angle $\theta$ ( $0 \leq \theta \leq \pi$ ): $\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$ , hence $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$ . Useful corollary: $\sin^2(\theta/2) = \dfrac{1-\cos\theta}{2} = \dfrac{|\vec{a}-\vec{b}|^2}{4|\vec{a}||\vec{b}|}$ when $\vec{a}, \vec{b}$ are unit vectors. The angle with an AXIS does not care about the axis vector's length: the angle between $\vec{a}$ and $2\hat{i}$ (or $-5\hat{i}$ , up to supplement) is the same as with $\hat{i}$ , $\cos\alpha = a_1/|\vec{a}|$ — the direction cosine. Scale factors cancel in the ratio, so never carry the 2 into the arithmetic.

Angle from dot product

\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}

$\theta$ angle between $\vec{a}$ and $\vec{b}$ , measured in $[0, \pi]$
$\vec{a}\cdot\vec{b}$ dot product (scalar)
$|\vec{a}|, |\vec{b}|$ magnitudes (always positive)

Visualization · project a onto b

a₁: 3a₂: 4b₁: 6b₂: 0

a·b = 18proj = (a·b)/|b| = 3.00θ ≈ 53°

The green band is how far a reaches along b — its signed length is the scalar projection (a·b)/|b|. Push the angle past 90° and the dot product, and the projection, turn negative.

Worked example

Find the angle between

\vec{a} = \hat{i} + \hat{j}

and

\vec{b} = \hat{i}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q95Moderate

Example 3 · Vectors · Dot Product and Angle

Consider vectors

\vec{a}=(0,1,1)

and

\vec{b}=(1,0,1)

. (1) Number of unit vectors perpendicular to both is only one. (2) Angle between them is

\pi/3

. Which is/are correct?

Obtuse angle iff $\vec{a}\cdot\vec{b} < 0$

Quadratic-in-

x

PYQs frequently ask for values of a parameter that make the angle obtuse. Set up the inequality

\vec{a}\cdot\vec{b} < 0

, solve as a quadratic, then exclude the boundary case

\vec{a}\cdot\vec{b} = 0

(which is perpendicular, not obtuse) AND values that make the vectors antiparallel (then

\theta = \pi

, the obtuse extreme).

Direction matters when comparing two angles

The angle between

\vec{a}

and

-\vec{a}

\pi

, not 0. If a problem asks for the angle between

\vec{a}

and

\vec{a} - \vec{b}

, don't carelessly subtract magnitudes — use the formula end-to-end.

Drill 8 more on angle between two vectors via the dot-product formula

Concept 4 of 5: Solving for an angle from a perpendicularity / magnitude constraint

Many MODERATE PYQs hand you a constraint like

(\vec{a}+2\vec{b}) \perp (5\vec{a}-4\vec{b})

|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|=k

and ask for the angle between

\vec{a}

and

\vec{b}

. The workflow is always the same: expand the constraint as a dot product, collect terms in

|\vec{a}|^2, |\vec{b}|^2

and

\vec{a}\cdot\vec{b}

, substitute the given magnitudes (often both 1 for unit vectors), and solve a single linear equation for

\cos\theta

Definition

Given a perpendicularity constraint $(\alpha\vec{a}+\beta\vec{b})\cdot(\gamma\vec{a}+\delta\vec{b}) = 0$ , expand using distributivity: $\alpha\gamma|\vec{a}|^2 + (\alpha\delta + \beta\gamma)\,\vec{a}\cdot\vec{b} + \beta\delta|\vec{b}|^2 = 0$ . Substitute known magnitudes and isolate $\vec{a}\cdot\vec{b}$ , then plug into the angle formula.

Expansion template

(\alpha\vec{a}+\beta\vec{b})\cdot(\gamma\vec{a}+\delta\vec{b}) = \alpha\gamma|\vec{a}|^2 + (\alpha\delta+\beta\gamma)\,\vec{a}\cdot\vec{b} + \beta\delta|\vec{b}|^2

$\alpha, \beta, \gamma, \delta$ given scalar coefficients
$|\vec{a}|, |\vec{b}|$ given (often $=1$ for unit vectors)
$\vec{a}\cdot\vec{b}$ unknown — solve for it, then read off $\theta$

Worked example

Let

\vec{a}, \vec{b}

be unit vectors such that

(2\vec{a}+\vec{b}) \perp (\vec{a}-2\vec{b})

. Find the angle between

\vec{a}

and

\vec{b}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2022 · Q67Moderate

Example 4 · Vectors · Dot Product and Angle

Let

\vec{a}

and

\vec{b}

are two unit vectors such that

\vec{a}+2\vec{b}

and

5\vec{a}-4\vec{b}

are perpendicular. What is the angle between

\vec{a}

and

\vec{b}

Don't forget the cross terms when expanding

(\vec{a}+2\vec{b})\cdot(5\vec{a}-4\vec{b})

has four products, not two — there are two

\vec{a}\cdot\vec{b}

terms that combine into the coefficient

\alpha\delta + \beta\gamma

. A factor-of-2 distractor often results from dropping one of them.

Unit vectors mean $|\vec{a}|^2 = 1$ , not $\vec{a} = 1$

When the magnitudes are stated as 1, the

|\vec{a}|^2

and

|\vec{b}|^2

terms simplify to 1, NOT zero. Some students drop them by analogy with

\vec{a}\cdot\vec{a}

when

\vec{a}

is the zero vector — wrong.

Drill 3 more on solving for an angle from a perpendicularity / magnitude constraint

Concept 5 of 5: Unit vectors, orthogonal triples, and decomposition

Three mutually-perpendicular unit vectors form an orthonormal basis — pairwise dot products are zero, self dot products are one. Whenever a problem says

\vec{a}, \vec{b}, \vec{c}

are unit and mutually perpendicular, you can compute any linear combination's magnitude or dot in two lines. Decomposing a vector along three given directions reduces to a small linear system once you take dot products with each direction.

Definition

Vectors $\vec{a}, \vec{b}, \vec{c}$ form an orthonormal triple if $\vec{a}\cdot\vec{a} = \vec{b}\cdot\vec{b} = \vec{c}\cdot\vec{c} = 1$ and $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$ . If $\vec{c} = p\vec{a} + q\vec{b} + r(\vec{a}\times\vec{b})$ with $\{\vec{a},\vec{b},\vec{a}\times\vec{b}\}$ orthonormal, the coefficients drop out as $p = \vec{c}\cdot\vec{a}$ , $q = \vec{c}\cdot\vec{b}$ , $r^2 = |\vec{c}|^2 - p^2 - q^2$ .

Orthonormal-triple identities

\vec{a}\cdot\vec{a} = 1, \quad \vec{a}\cdot\vec{b} = 0 \;\;(\text{if } \vec{a}\neq\vec{b}) \quad \text{for an orthonormal triple}

$\vec{a}, \vec{b}, \vec{c}$ three mutually-perpendicular unit vectors
$p, q, r$ decomposition coefficients along $\vec{a}, \vec{b}, \vec{a}\times\vec{b}$

Diagram · orthonormal triple î, ĵ, k̂ (drag to rotate)

î, ĵ, k̂ are mutually perpendicular unit vectors: each pair has dot product 0 (î·ĵ = ĵ·k̂ = k̂·î = 0) and each has length 1. They form the standard basis — any vector is a unique combination x î + y ĵ + z k̂.

Worked example

Let

\vec{a}, \vec{b}, \vec{c}

be three mutually perpendicular unit vectors. Find

|2\vec{a} - \vec{b} + 2\vec{c}|

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2018 · Q67Easy

Example 5 · Vectors · Dot Product and Angle

Let

\vec{a}, \vec{b}

and

\vec{c}

be three mutually perpendicular vectors each of unit magnitude. If

\vec{A}=\vec{a}+\vec{b}+\vec{c}

\vec{B}=\vec{a}-\vec{b}+\vec{c}

and

\vec{C}=\vec{a}-\vec{b}-\vec{c}

, then which one of the following is correct?

Three unit vectors at equal pairwise angles need not be orthonormal

\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = k

, the triple is symmetric but only orthonormal when

k = 0

. For other

k

values (e.g.

k = -1/2

— three coplanar vectors at

120^\circ

) the magnitudes of linear combinations look quite different.

$\{\vec{a}, \vec{b}, \vec{a}\times\vec{b}\}$ is orthonormal iff $\vec{a}\perp\vec{b}$ and both unit

\vec{a}, \vec{b}

are unit and perpendicular, then

|\vec{a}\times\vec{b}| = \sin 90^\circ = 1

— so the triple is orthonormal. If

\vec{a}\cdot\vec{b} \neq 0

, the cross product still produces a perpendicular vector, but it's not a unit vector and the basis isn't orthonormal.

Drill 5 more on unit vectors, orthogonal triples, and decomposition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Dot product — components form and work done
Dot product (components form)
$\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad W = \vec{F}\cdot\vec{d}$
Perpendicularity Test
Equivalent perpendicularity statements
$\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$
Angle between two vectors via the dot-product formula
Angle from dot product
$\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$
Solving for an angle from a perpendicularity / magnitude constraint
Expansion template
$(\alpha\vec{a}+\beta\vec{b})\cdot(\gamma\vec{a}+\delta\vec{b}) = \alpha\gamma|\vec{a}|^2 + (\alpha\delta+\beta\gamma)\,\vec{a}\cdot\vec{b} + \beta\delta|\vec{b}|^2$
Unit vectors, orthogonal triples, and decomposition
Orthonormal-triple identities
$\vec{a}\cdot\vec{a} = 1, \quad \vec{a}\cdot\vec{b} = 0 \;\;(\text{if } \vec{a}\neq\vec{b}) \quad \text{for an orthonormal triple}$

Watch out for (10)

Dot product gives a scalar; cross product gives a vector
→ Dot product — components form and work done
Work done is signed — negative work is fine
→ Dot product — components form and work done
$|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$ means $\vec{a}\perp\vec{b}$ , not $\vec{a}=\vec{b}$
→ Perpendicularity Test
Zero dot product needs both vectors non-zero
→ Perpendicularity Test
Obtuse angle iff $\vec{a}\cdot\vec{b} < 0$
→ Angle between two vectors via the dot-product formula
Direction matters when comparing two angles
→ Angle between two vectors via the dot-product formula
Don't forget the cross terms when expanding
→ Solving for an angle from a perpendicularity / magnitude constraint
Unit vectors mean $|\vec{a}|^2 = 1$ , not $\vec{a} = 1$
→ Solving for an angle from a perpendicularity / magnitude constraint
Three unit vectors at equal pairwise angles need not be orthonormal
→ Unit vectors, orthogonal triples, and decomposition
$\{\vec{a}, \vec{b}, \vec{a}\times\vec{b}\}$ is orthonormal iff $\vec{a}\perp\vec{b}$ and both unit
→ Unit vectors, orthogonal triples, and decomposition

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