8 min readConcept sheet1-mark MCQ and assertion-reason; part of 2-mark questions

Types of vectors

Zero, unit, coinitial, collinear, equal and negative vectors differ in length, direction or starting point. Equal vectors need the same magnitude and direction, wherever they start.

Zero vector 0⃗Magnitude 0; initial and terminal points coincide (AA⃗); its direction is not defined

Unit vectorMagnitude 1; â = a⃗/|a⃗| is the unit vector along a non-zero a⃗

Collinear (parallel) vectorsParallel to the same line, whatever their magnitudes and whether they point the same way or opposite ways

Equal vectorsSame magnitude and same direction, whatever their initial points

The types at a glance

Type	Meaning	Example
Zero (null) vector	Magnitude 0, written 0⃗	AA⃗; displacement of someone who returns home
Unit vector	Magnitude 1	î, ĵ, k̂ and (1/√2)(î + ĵ)
Coinitial vectors	Same initial point	AB⃗ and AC⃗
Collinear (parallel) vectors	Parallel to the same line	2î and −5î
Equal vectors	Same magnitude and direction	AB⃗ and DC⃗ in a parallelogram ABCD
Negative of a vector	Same magnitude, opposite direction	BA⃗ = −AB⃗

Collinear vs equal vectors

CollinearOnly need to be parallel to one line. They may have different lengths and may point opposite ways.

EqualMust have the same length and the same direction. Every pair of equal vectors is collinear, but not the other way round.

Test: non-zero b⃗ is collinear with a⃗ when b⃗ = λa⃗ for some scalar λ. They are equal only when λ = 1.

Free vectors

In this chapter, vectors are free vectors: you may slide a vector parallel to itself without changing it.
So in a parallelogram ABCD, AB⃗ = DC⃗ even though they start at different points.
Two vectors in component form are equal exactly when their î, ĵ and k̂ parts are equal.

Example 1: vectors in a square

Problem

ABCD is a square. Name (i) two equal vectors, (ii) two collinear vectors that are not equal, (iii) two coinitial vectors, (iv) a vector equal to −AD⃗.

Working

In a square, opposite sides are parallel and equal in length.
AB⃗ and DC⃗ have the same length and point the same way, so AB⃗ = DC⃗.
AB⃗ and CD⃗ are parallel but point opposite ways, so they are collinear but not equal.
AB⃗ and AD⃗ both start at A, so they are coinitial.
−AD⃗ = DA⃗, and DA⃗ = CB⃗ because DA and CB are opposite sides pointing the same way.

Answer

(i) AB⃗ = DC⃗; (ii) AB⃗ and CD⃗; (iii) AB⃗ and AD⃗; (iv) DA⃗, which equals CB⃗.

Example 2: is it a unit vector?

Problem

Check whether a⃗ = (1/3)î + (2/3)ĵ + (2/3)k̂ and b⃗ = î + ĵ + k̂ are unit vectors. If not, find the unit vector along it.

Working

|a⃗| = √(1/9 + 4/9 + 4/9) = √(9/9) = 1, so a⃗ is a unit vector.
|b⃗| = √(1 + 1 + 1) = √3, which is not 1.
Unit vector along b⃗ = b⃗/|b⃗| = (1/√3)(î + ĵ + k̂).

Answer

a⃗ is a unit vector; b⃗ is not. The unit vector along b⃗ is (1/√3)(î + ĵ + k̂).

In real life

Two trains on the same railway line, one going east and one going west, have collinear velocity vectors. Two trains on parallel tracks going east at the same speed have equal velocity vectors.

Exam angle

1 marks: “Are the vectors 3î − 2ĵ and −6î + 4ĵ collinear? Give a reason.” Yes. −6î + 4ĵ = −2(3î − 2ĵ), so one is a scalar multiple of the other. They point in opposite directions.

1 marks: “Write the unit vector along a⃗ = 3î − 4ĵ.” |a⃗| = √(9 + 16) = 5, so â = (3î − 4ĵ)/5 = (3/5)î − (4/5)ĵ.

2 marks: “Find x and y if 2xî + 3ĵ = 8î + (y − 1)ĵ.” Equal vectors have equal components: 2x = 8 and 3 = y − 1. So x = 4 and y = 4.

Remember it

Equal = same length + same way. Collinear = same line only.