12 min readConcept sheet1-mark MCQ or assertion-reason; 2-mark and 3-mark questions on angle, projection and perpendicular vectors; often a case-based question; also a JEE favourite

Scalar (dot) product

a⃗ · b⃗ = |a⃗| |b⃗| cos θ is a scalar. It is zero for perpendicular vectors, and it gives the angle and the projection of one vector on another.

Definitiona⃗ · b⃗ = |a⃗| |b⃗| cos θ, where θ (0 ≤ θ ≤ π) is the angle between a⃗ and b⃗

In componentsa⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃

Perpendicular testFor non-zero a⃗, b⃗: a⃗ · b⃗ = 0 exactly when a⃗ ⊥ b⃗

Projection of a⃗ on b⃗a⃗ · b̂ = (a⃗ · b⃗)/|b⃗|

Definition and meaning

Place a⃗ and b⃗ tail to tail and let θ be the angle between them. Their scalar (dot) product is a⃗ · b⃗ = |a⃗| |b⃗| cos θ. The result is a real number, not a vector. It is positive when θ is acute, negative when θ is obtuse, and zero when θ = 90°. Since î, ĵ, k̂ are perpendicular unit vectors, î · î = ĵ · ĵ = k̂ · k̂ = 1 and î · ĵ = ĵ · k̂ = k̂ · î = 0. This gives the component formula a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃.

Properties

Commutative: a⃗ · b⃗ = b⃗ · a⃗.
Distributive: a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗.
(λa⃗) · b⃗ = λ(a⃗ · b⃗) = a⃗ · (λb⃗).
a⃗ · a⃗ = |a⃗|², so |a⃗| = √(a⃗ · a⃗).
Angle: cos θ = (a⃗ · b⃗)/(|a⃗| |b⃗|).
θ = 0 gives a⃗ · b⃗ = |a⃗| |b⃗|; θ = π gives a⃗ · b⃗ = −|a⃗| |b⃗|.
|a⃗ ± b⃗|² = |a⃗|² ± 2a⃗ · b⃗ + |b⃗|², and (a⃗ + b⃗) · (a⃗ − b⃗) = |a⃗|² − |b⃗|².

Geometrical meaning: projection (and how to draw it)

1
Draw OA⃗ = a⃗ and OB⃗ = b⃗ from the same point O, with angle θ between them.
2
From A, drop a perpendicular to the line OB. Call its foot M and mark the right angle.
3
OM = |a⃗| cos θ is the projection of a⃗ on b⃗. Label it.
4
Since a⃗ · b⃗ = |a⃗| |b⃗| cos θ, the projection = (a⃗ · b⃗)/|b⃗| = a⃗ · b̂.
5
It is positive for acute θ and negative for obtuse θ. The projection vector is (a⃗ · b̂) b̂.

Example 1: angle between two vectors

Problem

Find the angle between a⃗ = 2î + 2ĵ − k̂ and b⃗ = 6î − 3ĵ + 2k̂.

Working

a⃗ · b⃗ = (2)(6) + (2)(−3) + (−1)(2) = 12 − 6 − 2 = 4
|a⃗| = √(4 + 4 + 1) = 3, |b⃗| = √(36 + 9 + 4) = 7
cos θ = (a⃗ · b⃗)/(|a⃗| |b⃗|) = 4/21
θ = cos⁻¹(4/21)

Answer

θ = cos⁻¹(4/21)

Example 2: perpendicular vectors and projection

Problem

(i) Find λ if 3î + λĵ − 2k̂ and 2î − 4ĵ + k̂ are perpendicular. (ii) Find the projection of 4î − ĵ + 3k̂ on 2î + ĵ − 2k̂.

Working

(i) Perpendicular means the dot product is 0.
(3)(2) + (λ)(−4) + (−2)(1) = 0, so 6 − 4λ − 2 = 0.
4λ = 4, so λ = 1.
(ii) a⃗ · b⃗ = 8 − 1 − 6 = 1 and |b⃗| = √(4 + 1 + 4) = 3.
Projection = (a⃗ · b⃗)/|b⃗| = 1/3.

Answer

(i) λ = 1; (ii) projection = 1/3.

Example 3: board-style

Problem

a⃗ + b⃗ + c⃗ = 0⃗, with |a⃗| = 2, |b⃗| = 3 and |c⃗| = √19. Find the angle between a⃗ and b⃗.

Working

a⃗ + b⃗ = −c⃗, so |a⃗ + b⃗|² = |c⃗|².
|a⃗|² + 2a⃗ · b⃗ + |b⃗|² = |c⃗|²
4 + 2a⃗ · b⃗ + 9 = 19, so a⃗ · b⃗ = 3.
cos θ = 3/(2 × 3) = 1/2
θ = π/3

Answer

The angle between a⃗ and b⃗ is π/3 (60°).

In real life

Work done by a force is a dot product, W = F⃗ · d⃗. When you pull a suitcase by a slanted handle, only the part of your force along the floor does work.

Exam angle

1 marks: “If |a⃗| = 2, |b⃗| = 3 and a⃗ · b⃗ = 3, find the angle between a⃗ and b⃗.” cos θ = 3/(2 × 3) = 1/2, so θ = π/3.

2 marks: “Find the projection of a⃗ = î + 2ĵ + 2k̂ on b⃗ = 3î + 4k̂.” a⃗ · b⃗ = 3 + 0 + 8 = 11 and |b⃗| = 5. Projection = 11/5.

3 marks: “a⃗ + b⃗ + c⃗ = 0⃗ with |a⃗| = 2, |b⃗| = 4, |c⃗| = 5. Find a⃗ · b⃗ + b⃗ · c⃗ + c⃗ · a⃗.” |a⃗ + b⃗ + c⃗|² = 0, so |a⃗|² + |b⃗|² + |c⃗|² + 2(a⃗ · b⃗ + b⃗ · c⃗ + c⃗ · a⃗) = 0. 4 + 16 + 25 + 2S = 0, so S = −45/2.

Remember it

Dot gives a number; a zero dot means a right angle.