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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.

All Vectors notesNDA Maths strategy

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.

Concept 1 of 5

Dot product — components form and work done

Intuition

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 F⃗\vec{F} displaces a particle by d⃗\vec{d}, the work done is exactly F⃗⋅d⃗\vec{F}\cdot\vec{d}.

Definition

If a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}, then a⃗⋅b⃗=a1b1+a2b2+a3b3\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=F⃗⋅d⃗W = \vec{F}\cdot\vec{d} where d⃗\vec{d} is the displacement.

Dot product (components form)

a⃗⋅b⃗=a1b1+a2b2+a3b3W=F⃗⋅d⃗\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad W = \vec{F}\cdot\vec{d}
  • ai,bia_i, b_icomponents of a⃗,b⃗\vec{a}, \vec{b} along i^,j^,k^\hat{i}, \hat{j}, \hat{k}
  • WWwork done by a constant force F⃗\vec{F} through displacement d⃗\vec{d}

Diagram · work = F · d = |F||d| cos θ

θFdF 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 F⃗=2i^+j^−k^\vec{F} = 2\hat{i} + \hat{j} - \hat{k} displaces a particle from A(1,0,2)A(1, 0, 2) to B(3,2,1)B(3, 2, 1). Find the work done.
  1. Compute the displacement: d⃗=AB→=(3−1)i^+(2−0)j^+(1−2)k^=2i^+2j^−k^\vec{d} = \overrightarrow{AB} = (3-1)\hat{i} + (2-0)\hat{j} + (1-2)\hat{k} = 2\hat{i} + 2\hat{j} - \hat{k}.
  2. Apply W=F⃗⋅d⃗=2⋅2+1⋅2+(−1)⋅(−1)=4+2+1W = \vec{F}\cdot\vec{d} = 2\cdot 2 + 1\cdot 2 + (-1)\cdot(-1) = 4 + 2 + 1.
  3. Sum: W=7W = 7.
Answer:W=7W = 7 units of work
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate a⃗⋅b⃗\vec{a}\cdot\vec{b} for a⃗=2i^−j^+3k^\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} and b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    a⃗⋅b⃗\vec{a}\cdot\vec{b} for a⃗=i^+2j^\vec{a} = \hat{i} + 2\hat{j}, b⃗=3i^+j^\vec{b} = 3\hat{i} + \hat{j}?
  2. 2.
    i^⋅j^=?\hat{i}\cdot\hat{j} = ?
  3. 3.
    i^⋅i^=?\hat{i}\cdot\hat{i} = ?
  4. 4.
    a⃗⋅b⃗\vec{a}\cdot\vec{b} for a⃗=2i^−j^+k^\vec{a} = 2\hat{i} - \hat{j} + \hat{k}, b⃗=i^+j^−k^\vec{b} = \hat{i} + \hat{j} - \hat{k}?

From the bank · past-year question

Example 1VectorsEASY
A force F⃗=i^+3j^+2k^\vec{F} = \hat{i}+3\hat{j}+2\hat{k} acts on a particle to displace it from the point A(i^+2j^−3k^)A(\hat{i}+2\hat{j}-3\hat{k}) to the point B(3i^−j^+5k^)B(3\hat{i}-\hat{j}+5\hat{k}). The work done by the force will be

[Q47 · Sep · 2017]

Dot product gives a scalar; cross product gives a vector

An MCQ option that returns a vector for a⃗⋅b⃗\vec{a}\cdot\vec{b}, or a scalar for a⃗×b⃗\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

Intuition

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: a⃗⋅b⃗=0\vec{a}\cdot\vec{b}=0 directly; ∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}| (the diagonals of a parallelogram are equal iff it's a rectangle); and (a⃗+b⃗)⋅(a⃗+b⃗)=∣a⃗∣2+∣b⃗∣2(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 (the Pythagoras identity).

Definition

For non-zero a⃗,b⃗\vec{a}, \vec{b}, all three of the following are equivalent: a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0; ∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|; (a⃗+b⃗)⋅(a⃗+b⃗)=∣a⃗∣2+∣b⃗∣2(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2. Each says a⃗⊥b⃗\vec{a} \perp \vec{b}.

Equivalent perpendicularity statements

a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0  ⟺  ∣a⃗+b⃗∣=∣a⃗−b⃗∣\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|
  • a⃗⋅b⃗\vec{a}\cdot\vec{b}scalar dot product
  • ∣a⃗±b⃗∣|\vec{a}\pm\vec{b}|magnitudes of the diagonals of the parallelogram on a⃗,b⃗\vec{a}, \vec{b}

Worked example

Find the value of λ\lambda so that a⃗=2i^−3j^+k^\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k} and b⃗=i^+λj^−5k^\vec{b} = \hat{i} + \lambda\hat{j} - 5\hat{k} are perpendicular.
  1. Apply the perpendicularity test a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0.
  2. Compute the dot: 2⋅1+(−3)⋅λ+1⋅(−5)=2−3λ−5=−3−3λ2\cdot 1 + (-3)\cdot\lambda + 1\cdot(-5) = 2 - 3\lambda - 5 = -3 - 3\lambda.
  3. Set equal to zero: −3−3λ=0  ⇒  λ=−1-3 - 3\lambda = 0 \;\Rightarrow\; \lambda = -1.
Answer:λ=−1\lambda = -1
Practice this conceptself-check · 4 quick reps

Try it yourself

Find λ\lambda so that a⃗=3i^+2j^+λk^\vec{a} = 3\hat{i} + 2\hat{j} + \lambda\hat{k} and b⃗=i^−4j^+2k^\vec{b} = \hat{i} - 4\hat{j} + 2\hat{k} are perpendicular.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Are i^+j^\hat{i} + \hat{j} and i^−j^\hat{i} - \hat{j} perpendicular?
  2. 2.
    λ\lambda for a⃗=i^+λj^\vec{a} = \hat{i} + \lambda\hat{j}, b⃗=2i^−j^\vec{b} = 2\hat{i} - \hat{j} perpendicular?
  3. 3.
    Perpendicularity condition for a⃗,b⃗\vec{a}, \vec{b}?
  4. 4.
    Are 2i^+3j^2\hat{i} + 3\hat{j} and 3i^−2j^3\hat{i} - 2\hat{j} perpendicular?

From the bank · past-year question

Example 2VectorsEASY
What is the value of λ\lambda for which the vectors 3i^+4j^−k^3\hat{i}+4\hat{j}-\hat{k} and −2i^+λj^+10k^-2\hat{i}+\lambda\hat{j}+10\hat{k} are perpendicular?

[Q70 · Apr · 2019]

∣a⃗+b⃗∣=∣a⃗−b⃗∣|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}| means a⃗⊥b⃗\vec{a}\perp\vec{b}, not a⃗=b⃗\vec{a}=\vec{b}

Square both sides: ∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2=∣a⃗∣2−2a⃗⋅b⃗+∣b⃗∣2|\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 4a⃗⋅b⃗=04\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 0⃗⋅a⃗=0\vec{0}\cdot\vec{a} = 0 for any a⃗\vec{a}, but 0⃗\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

Intuition

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>0) versus obtuse (<0<0); =0=0 is perpendicular, the special case above.

Definition

For non-zero a⃗,b⃗\vec{a}, \vec{b} at angle θ\theta (0≤θ≤π0 \leq \theta \leq \pi): a⃗⋅b⃗=∣a⃗∣ ∣b⃗∣cos⁡θ\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta, hence cos⁡θ=a⃗⋅b⃗∣a⃗∣ ∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}. Useful corollary: sin⁡2(θ/2)=1−cos⁡θ2=∣a⃗−b⃗∣24∣a⃗∣∣b⃗∣\sin^2(\theta/2) = \dfrac{1-\cos\theta}{2} = \dfrac{|\vec{a}-\vec{b}|^2}{4|\vec{a}||\vec{b}|} when a⃗,b⃗\vec{a}, \vec{b} are unit vectors.

Angle from dot product

cos⁡θ=a⃗⋅b⃗∣a⃗∣ ∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}
  • θ\thetaangle between a⃗\vec{a} and b⃗\vec{b}, measured in [0,π][0, \pi]
  • a⃗⋅b⃗\vec{a}\cdot\vec{b}dot product (scalar)
  • ∣a⃗∣,∣b⃗∣|\vec{a}|, |\vec{b}|magnitudes (always positive)

Visualization · project a onto b

ab
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 a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=i^\vec{b} = \hat{i}.
  1. Dot product: a⃗⋅b⃗=1⋅1+1⋅0=1\vec{a}\cdot\vec{b} = 1\cdot 1 + 1\cdot 0 = 1.
  2. Magnitudes: ∣a⃗∣=1+1=2|\vec{a}| = \sqrt{1+1} = \sqrt{2}; ∣b⃗∣=1|\vec{b}| = 1.
  3. Apply the formula: cos⁡θ=12⋅1=12\cos\theta = \dfrac{1}{\sqrt{2}\cdot 1} = \dfrac{1}{\sqrt{2}}.
  4. Hence θ=π4=45∘\theta = \dfrac{\pi}{4} = 45^\circ.
Answer:θ=π4\theta = \dfrac{\pi}{4} (i.e. 45∘45^\circ)
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the angle between a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=i^−j^\vec{b} = \hat{i} - \hat{j}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Angle between i^\hat{i} and j^\hat{j}?
  2. 2.
    cos⁡θ\cos\theta for a⃗⋅b⃗=6\vec{a}\cdot\vec{b} = 6, ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=4|\vec{b}| = 4?
  3. 3.
    Angle between a⃗\vec{a} and a⃗\vec{a}?
  4. 4.
    If cos⁡θ<0\cos\theta < 0, the angle is?

From the bank · past-year question

Example 3VectorsMODERATE
Consider vectors a⃗=(0,1,1)\vec{a}=(0,1,1) and b⃗=(1,0,1)\vec{b}=(1,0,1). (1) Number of unit vectors perpendicular to both is only one. (2) Angle between them is π/3\pi/3. Which is/are correct?

[Q95 · Sep · 2023]

Obtuse angle iff a⃗⋅b⃗<0\vec{a}\cdot\vec{b} < 0

Quadratic-in-xx PYQs frequently ask for values of a parameter that make the angle obtuse. Set up the inequality a⃗⋅b⃗<0\vec{a}\cdot\vec{b} < 0, solve as a quadratic, then exclude the boundary case a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0 (which is perpendicular, not obtuse) AND values that make the vectors antiparallel (then θ=π\theta = \pi, the obtuse extreme).

Direction matters when comparing two angles

The angle between a⃗\vec{a} and −a⃗-\vec{a} is π\pi, not 0. If a problem asks for the angle between a⃗\vec{a} and a⃗−b⃗\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

Intuition

Many MODERATE PYQs hand you a constraint like (a⃗+2b⃗)⊥(5a⃗−4b⃗)(\vec{a}+2\vec{b}) \perp (5\vec{a}-4\vec{b}) or ∣a⃗+b⃗∣=∣a⃗−b⃗∣=k|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|=k and ask for the angle between a⃗\vec{a} and b⃗\vec{b}. The workflow is always the same: expand the constraint as a dot product, collect terms in ∣a⃗∣2,∣b⃗∣2|\vec{a}|^2, |\vec{b}|^2 and a⃗⋅b⃗\vec{a}\cdot\vec{b}, substitute the given magnitudes (often both 1 for unit vectors), and solve a single linear equation for cos⁡θ\cos\theta.

Definition

Given a perpendicularity constraint (αa⃗+βb⃗)⋅(γa⃗+δb⃗)=0(\alpha\vec{a}+\beta\vec{b})\cdot(\gamma\vec{a}+\delta\vec{b}) = 0, expand using distributivity: αγ∣a⃗∣2+(αδ+βγ) a⃗⋅b⃗+βδ∣b⃗∣2=0\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 a⃗⋅b⃗\vec{a}\cdot\vec{b}, then plug into the angle formula.

Expansion template

(αa⃗+βb⃗)⋅(γa⃗+δb⃗)=αγ∣a⃗∣2+(αδ+βγ) a⃗⋅b⃗+βδ∣b⃗∣2(\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
  • α,β,γ,δ\alpha, \beta, \gamma, \deltagiven scalar coefficients
  • ∣a⃗∣,∣b⃗∣|\vec{a}|, |\vec{b}|given (often =1=1 for unit vectors)
  • a⃗⋅b⃗\vec{a}\cdot\vec{b}unknown — solve for it, then read off θ\theta

Worked example

Let a⃗,b⃗\vec{a}, \vec{b} be unit vectors such that (2a⃗+b⃗)⊥(a⃗−2b⃗)(2\vec{a}+\vec{b}) \perp (\vec{a}-2\vec{b}). Find the angle between a⃗\vec{a} and b⃗\vec{b}.
  1. Set (2a⃗+b⃗)⋅(a⃗−2b⃗)=0(2\vec{a}+\vec{b})\cdot(\vec{a}-2\vec{b}) = 0.
  2. Expand: 2∣a⃗∣2+(−4+1) a⃗⋅b⃗−2∣b⃗∣2=2−3a⃗⋅b⃗−22|\vec{a}|^2 + (-4 + 1)\,\vec{a}\cdot\vec{b} - 2|\vec{b}|^2 = 2 - 3\vec{a}\cdot\vec{b} - 2 (using ∣a⃗∣=∣b⃗∣=1|\vec{a}|=|\vec{b}|=1).
  3. Set equal to zero: −3a⃗⋅b⃗=0  ⇒  a⃗⋅b⃗=0-3\vec{a}\cdot\vec{b} = 0 \;\Rightarrow\; \vec{a}\cdot\vec{b} = 0.
  4. Unit vectors with a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0: cos⁡θ=0\cos\theta = 0, so θ=π2\theta = \dfrac{\pi}{2}.
Answer:θ=π2\theta = \dfrac{\pi}{2} (i.e. 90∘90^\circ)
Practice this conceptself-check · 4 quick reps

Try it yourself

a⃗\vec{a} and b⃗\vec{b} are unit vectors with ∣a⃗−b⃗∣=1|\vec{a} - \vec{b}| = 1. Find the angle between them.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Expand (a⃗+b⃗)⋅(a⃗+b⃗)(\vec{a} + \vec{b})\cdot(\vec{a} + \vec{b}).
  2. 2.
    Unit a⃗,b⃗\vec{a}, \vec{b}: write ∣a⃗−b⃗∣2|\vec{a} - \vec{b}|^2 in terms of a⃗⋅b⃗\vec{a}\cdot\vec{b}.
  3. 3.
    Unit a⃗,b⃗\vec{a}, \vec{b} with ∣a⃗+b⃗∣2=3|\vec{a} + \vec{b}|^2 = 3. Find a⃗⋅b⃗\vec{a}\cdot\vec{b}.
  4. 4.
    For a unit vector, a⃗⋅a⃗=?\vec{a}\cdot\vec{a} = ?

From the bank · past-year question

Example 4VectorsMODERATE
Let a⃗\vec{a} and b⃗\vec{b} are two unit vectors such that a⃗+2b⃗\vec{a}+2\vec{b} and 5a⃗−4b⃗5\vec{a}-4\vec{b} are perpendicular. What is the angle between a⃗\vec{a} and b⃗\vec{b}?

[Q67 · Sep · 2022]

Don't forget the cross terms when expanding

(a⃗+2b⃗)⋅(5a⃗−4b⃗)(\vec{a}+2\vec{b})\cdot(5\vec{a}-4\vec{b}) has four products, not two — there are two a⃗⋅b⃗\vec{a}\cdot\vec{b} terms that combine into the coefficient αδ+βγ\alpha\delta + \beta\gamma. A factor-of-2 distractor often results from dropping one of them.

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

When the magnitudes are stated as 1, the ∣a⃗∣2|\vec{a}|^2 and ∣b⃗∣2|\vec{b}|^2 terms simplify to 1, NOT zero. Some students drop them by analogy with a⃗⋅a⃗\vec{a}\cdot\vec{a} when a⃗\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

Intuition

Three mutually-perpendicular unit vectors form an orthonormal basis — pairwise dot products are zero, self dot products are one. Whenever a problem says a⃗,b⃗,c⃗\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 a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} form an orthonormal triple if a⃗⋅a⃗=b⃗⋅b⃗=c⃗⋅c⃗=1\vec{a}\cdot\vec{a} = \vec{b}\cdot\vec{b} = \vec{c}\cdot\vec{c} = 1 and a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=0\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0. If c⃗=pa⃗+qb⃗+r(a⃗×b⃗)\vec{c} = p\vec{a} + q\vec{b} + r(\vec{a}\times\vec{b}) with {a⃗,b⃗,a⃗×b⃗}\{\vec{a},\vec{b},\vec{a}\times\vec{b}\} orthonormal, the coefficients drop out as p=c⃗⋅a⃗p = \vec{c}\cdot\vec{a}, q=c⃗⋅b⃗q = \vec{c}\cdot\vec{b}, r2=∣c⃗∣2−p2−q2r^2 = |\vec{c}|^2 - p^2 - q^2.

Orthonormal-triple identities

a⃗⋅a⃗=1,a⃗⋅b⃗=0    (if a⃗≠b⃗)for an orthonormal triple\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}
  • a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}three mutually-perpendicular unit vectors
  • p,q,rp, q, rdecomposition coefficients along a⃗,b⃗,a⃗×b⃗\vec{a}, \vec{b}, \vec{a}\times\vec{b}

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

îĵk̂

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 a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three mutually perpendicular unit vectors. Find ∣2a⃗−b⃗+2c⃗∣|2\vec{a} - \vec{b} + 2\vec{c}|.
  1. Expand the square: ∣2a⃗−b⃗+2c⃗∣2=4∣a⃗∣2+∣b⃗∣2+4∣c⃗∣2+(cross terms)|2\vec{a} - \vec{b} + 2\vec{c}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2 + \text{(cross terms)}.
  2. Self-dots: 4(1)+1(1)+4(1)=94(1) + 1(1) + 4(1) = 9. All pairwise products vanish by mutual perpendicularity.
  3. So ∣2a⃗−b⃗+2c⃗∣2=9|2\vec{a} - \vec{b} + 2\vec{c}|^2 = 9, giving ∣2a⃗−b⃗+2c⃗∣=3|2\vec{a} - \vec{b} + 2\vec{c}| = 3.
Answer:∣2a⃗−b⃗+2c⃗∣=3|2\vec{a} - \vec{b} + 2\vec{c}| = 3
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a⃗\vec{a} and b⃗\vec{b} are perpendicular unit vectors. Find ∣3a⃗−4b⃗∣|3\vec{a} - 4\vec{b}|.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} mutually perpendicular unit vectors. ∣a⃗+b⃗+c⃗∣|\vec{a} + \vec{b} + \vec{c}|?
  2. 2.
    Perpendicular unit a⃗,b⃗\vec{a}, \vec{b}: ∣a⃗+b⃗∣|\vec{a} + \vec{b}|?
  3. 3.
    Orthonormal triple: a⃗⋅b⃗=?\vec{a}\cdot\vec{b} = ?
  4. 4.
    Perpendicular unit a⃗,b⃗\vec{a}, \vec{b}: ∣3a⃗+4b⃗∣|3\vec{a} + 4\vec{b}|?

From the bank · past-year question

Example 5VectorsEASY
Let a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} be three mutually perpendicular vectors each of unit magnitude. If A⃗=a⃗+b⃗+c⃗\vec{A}=\vec{a}+\vec{b}+\vec{c}, B⃗=a⃗−b⃗+c⃗\vec{B}=\vec{a}-\vec{b}+\vec{c} and C⃗=a⃗−b⃗−c⃗\vec{C}=\vec{a}-\vec{b}-\vec{c}, then which one of the following is correct?

[Q67 · Sep · 2018]

Three unit vectors at equal pairwise angles need not be orthonormal

If a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=k\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=0k = 0. For other kk values (e.g. k=−1/2k = -1/2 — three coplanar vectors at 120∘120^\circ) the magnitudes of linear combinations look quite different.

{a⃗,b⃗,a⃗×b⃗}\{\vec{a}, \vec{b}, \vec{a}\times\vec{b}\} is orthonormal iff a⃗⊥b⃗\vec{a}\perp\vec{b} and both unit

If a⃗,b⃗\vec{a}, \vec{b} are unit and perpendicular, then ∣a⃗×b⃗∣=sin⁡90∘=1|\vec{a}\times\vec{b}| = \sin 90^\circ = 1 — so the triple is orthonormal. If a⃗⋅b⃗≠0\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)

    a⃗⋅b⃗=a1b1+a2b2+a3b3W=F⃗⋅d⃗\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

    a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0  ⟺  ∣a⃗+b⃗∣=∣a⃗−b⃗∣\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⁡θ=a⃗⋅b⃗∣a⃗∣ ∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}
  • Solving for an angle from a perpendicularity / magnitude constraint

    Expansion template

    (αa⃗+βb⃗)⋅(γa⃗+δb⃗)=αγ∣a⃗∣2+(αδ+βγ) a⃗⋅b⃗+βδ∣b⃗∣2(\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

    a⃗⋅a⃗=1,a⃗⋅b⃗=0    (if a⃗≠b⃗)for an orthonormal triple\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)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsEASY
If r⃗=xi^+yj^+zk^\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}, then what is r⃗⋅(i^+j^+k^)\vec{r}\cdot(\hat{i}+\hat{j}+\hat{k}) equal to?

[Q64 · Sep · 2018]

Example 2VectorsEASY
If a⃗=2i^+3j^+4k^\vec{a}=2\hat{i}+3\hat{j}+4\hat{k} and b⃗=3i^+2j^−λk^\vec{b}=3\hat{i}+2\hat{j}-\lambda\hat{k} are perpendicular, then what is the value of λ\lambda?

[Q70 · Apr · 2017]

Example 3VectorsMODERATE
For what value of the angle between the vectors a⃗\vec{a} and b⃗\vec{b} is the quantity ∣a⃗×b⃗∣+3∣a⃗⋅b⃗∣|\vec{a}\times\vec{b}|+\sqrt{3}|\vec{a}\cdot\vec{b}| maximum?

[Q52 · Sep · 2024]

Example 4VectorsMODERATE
Let θ\theta be the angle between two unit vectors a⃗\vec{a} and b⃗\vec{b}. If a⃗+2b⃗\vec{a}+2\vec{b} is perpendicular to 5a⃗−4b⃗5\vec{a}-4\vec{b}, then what is cos⁡θ+cos⁡2θ\cos\theta+\cos2\theta equal to?

[Q53 · Sep · 2024]

Example 5VectorsMODERATE
for the items that follow: Let a⃗\vec{a}, b⃗\vec{b}, c⃗\vec{c} be unit vectors. Further, a⃗\vec{a} is perpendicular to b⃗\vec{b}; c⃗\vec{c} makes angle π3\frac{\pi}{3} with both a⃗\vec{a} and b⃗\vec{b}; and c⃗=pa⃗+qb⃗+r(a⃗×b⃗)\vec{c}=p\vec{a}+q\vec{b}+r(\vec{a}\times\vec{b}).
What is the value of (p+q)(p+q)?

[Q69 · Apr · 2026]

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