9 min readConcept sheet1-mark MCQ; part of 2-mark and 3-mark questions

Addition of vectors

Place vectors head to tail: by the triangle law, a⃗ + b⃗ is the side that closes the triangle. Addition is commutative and associative.

Triangle lawAB⃗ + BC⃗ = AC⃗

Parallelogram lawIf OA⃗ = a⃗ and OB⃗ = b⃗ are adjacent sides of parallelogram OACB, then a⃗ + b⃗ = OC⃗, the diagonal through O

Propertiesa⃗ + b⃗ = b⃗ + a⃗; (a⃗ + b⃗) + c⃗ = a⃗ + (b⃗ + c⃗); a⃗ + 0⃗ = a⃗; a⃗ + (−a⃗) = 0⃗

Closed triangleAB⃗ + BC⃗ + CA⃗ = 0⃗

How to draw the triangle law

1
Draw a⃗ as an arrow from A to B.
2
From the head B, draw b⃗ as an arrow from B to C, with the same length and direction as b⃗.
3
Join A to C with an arrow from A to C. Label it a⃗ + b⃗.
4
Label the points A, B, C and all three arrows. The sum always runs from the first tail to the last head.

Parallelogram law and subtraction

If a⃗ and b⃗ start at the same point O, with OA⃗ = a⃗ and OB⃗ = b⃗, complete the parallelogram OACB. The diagonal OC⃗ = a⃗ + b⃗. This is the triangle law again, because AC⃗ = OB⃗ = b⃗. To subtract, add the negative: a⃗ − b⃗ = a⃗ + (−b⃗). In the same parallelogram, a⃗ − b⃗ is the other diagonal, BA⃗.

Properties of addition

Commutative: a⃗ + b⃗ = b⃗ + a⃗.
Associative: (a⃗ + b⃗) + c⃗ = a⃗ + (b⃗ + c⃗), so brackets can be dropped.
Additive identity: a⃗ + 0⃗ = a⃗.
Additive inverse: a⃗ + (−a⃗) = 0⃗.
Triangle inequality: |a⃗ + b⃗| ≤ |a⃗| + |b⃗|, because one side of a triangle is never longer than the other two together.
In components, add the matching parts: (a₁î + a₂ĵ + a₃k̂) + (b₁î + b₂ĵ + b₃k̂) = (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂.

Example 1: simplify a chain

Problem

Show that AB⃗ + BC⃗ + CD⃗ = AD⃗. Then simplify PQ⃗ + QR⃗ − PR⃗.

Working

Triangle law: AB⃗ + BC⃗ = AC⃗.
So AB⃗ + BC⃗ + CD⃗ = AC⃗ + CD⃗ = AD⃗.
Triangle law: PQ⃗ + QR⃗ = PR⃗.
So PQ⃗ + QR⃗ − PR⃗ = PR⃗ − PR⃗ = 0⃗.

Answer

AB⃗ + BC⃗ + CD⃗ = AD⃗, and PQ⃗ + QR⃗ − PR⃗ = 0⃗.

Example 2: in component form

Problem

a⃗ = 2î + ĵ − k̂ and b⃗ = î − 3ĵ + 4k̂. Find a⃗ + b⃗, its magnitude and the unit vector along it.

Working

a⃗ + b⃗ = (2 + 1)î + (1 − 3)ĵ + (−1 + 4)k̂
= 3î − 2ĵ + 3k̂
|a⃗ + b⃗| = √(9 + 4 + 9) = √22
Unit vector = (3î − 2ĵ + 3k̂)/√22

Answer

a⃗ + b⃗ = 3î − 2ĵ + 3k̂, magnitude √22, unit vector (3î − 2ĵ + 3k̂)/√22.

Example 3: crossing a river

Problem

A boat is rowed at 4 km/h straight across a river (towards the north). The current flows east at 3 km/h. Find the magnitude of the boat's actual velocity.

Working

Rowing velocity = 4ĵ, current = 3î (î east, ĵ north).
By the triangle law, actual velocity = 3î + 4ĵ.
Magnitude = √(3² + 4²) = √25 = 5 km/h.
Its direction makes an angle θ east of north with tan θ = 3/4.

Answer

5 km/h, at an angle tan⁻¹(3/4) east of north.

In real life

Row a boat across the Hooghly and the current pushes it downstream. Your real velocity is the vector sum of your rowing velocity and the current's velocity.

Exam angle

1 marks: “ABC is a triangle. Find AB⃗ + BC⃗ + CA⃗.” AB⃗ + BC⃗ = AC⃗, and AC⃗ + CA⃗ = 0⃗. So the sum is the zero vector.

2 marks: “Find the unit vector in the direction of the sum of a⃗ = 3î − ĵ + 2k̂ and b⃗ = −î + 3ĵ − k̂.” a⃗ + b⃗ = 2î + 2ĵ + k̂. Its magnitude is √(4 + 4 + 1) = 3. Unit vector = (2î + 2ĵ + k̂)/3.

Remember it

Head to tail, then close the triangle from the first tail to the last head.