9 min readConcept sheet1-mark MCQ; 2-mark questions on the vector joining two points and on collinear points

Components of a vector

Any vector can be written as xî + yĵ + zk̂. The vector from P₁ to P₂ is (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.

î, ĵ, k̂Unit vectors along the positive x, y and z axes

Component formr⃗ = xî + yĵ + zk̂; x, y, z are the scalar components and xî, yĵ, zk̂ the vector components

Magnitude|r⃗| = √(x² + y² + z²)

Vector joining P₁ to P₂P₁P₂⃗ = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂ (head minus tail)

Writing a vector in components

Let î, ĵ and k̂ be unit vectors along the x, y and z axes. For a point P(x, y, z), go x units along OX, then y units parallel to OY, then z units parallel to OZ. By the triangle law, OP⃗ = xî + yĵ + zk̂. This is the component form. The numbers x, y, z are the scalar components, and xî, yĵ, zk̂ are the vector components along the axes.

Working in components (a⃗ = a₁î + a₂ĵ + a₃k̂, b⃗ = b₁î + b₂ĵ + b₃k̂)

Sum: a⃗ + b⃗ = (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂.
Difference: a⃗ − b⃗ = (a₁ − b₁)î + (a₂ − b₂)ĵ + (a₃ − b₃)k̂.
Scalar multiple: λa⃗ = (λa₁)î + (λa₂)ĵ + (λa₃)k̂.
Equality: a⃗ = b⃗ exactly when a₁ = b₁, a₂ = b₂ and a₃ = b₃.
Collinear: b⃗ = λa⃗ exactly when b₁/a₁ = b₂/a₂ = b₃/a₃ = λ (when a₁, a₂, a₃ are non-zero).

Derivation: the vector joining two points

1
Let P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) have position vectors OP₁⃗ and OP₂⃗.
2
Triangle law: OP₁⃗ + P₁P₂⃗ = OP₂⃗.
3
So P₁P₂⃗ = OP₂⃗ − OP₁⃗.
4
= (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.
5
So, |P₁P₂⃗| = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²], the distance between the points.

Example 1: vector joining two points

Problem

Find the vector from P(1, −2, 3) to Q(3, 1, −3), its magnitude and the unit vector along it.

Working

PQ⃗ = (3 − 1)î + (1 − (−2))ĵ + (−3 − 3)k̂
= 2î + 3ĵ − 6k̂
|PQ⃗| = √(4 + 9 + 36) = √49 = 7
Unit vector = (2î + 3ĵ − 6k̂)/7

Answer

PQ⃗ = 2î + 3ĵ − 6k̂, |PQ⃗| = 7, unit vector (2/7)î + (3/7)ĵ − (6/7)k̂.

Example 2: board-style, collinear points

Problem

Show that the points A(1, −2, 3), B(2, 1, 1) and C(4, 7, −3) are collinear.

Working

AB⃗ = (2 − 1)î + (1 + 2)ĵ + (1 − 3)k̂ = î + 3ĵ − 2k̂
BC⃗ = (4 − 2)î + (7 − 1)ĵ + (−3 − 1)k̂ = 2î + 6ĵ − 4k̂
BC⃗ = 2(î + 3ĵ − 2k̂) = 2AB⃗, so AB⃗ and BC⃗ are parallel.
They share the point B, so A, B and C lie on one line.

Answer

BC⃗ = 2AB⃗ with B common, so A, B, C are collinear.

In real life

The GPS in a phone records your movement as east, north and up components. It then combines them to show how far you have travelled and in which direction.

Exam angle

1 marks: “Write the scalar and vector components of the vector joining A(1, 2, 3) to B(4, 6, 8).” AB⃗ = 3î + 4ĵ + 5k̂. Scalar components: 3, 4, 5. Vector components: 3î, 4ĵ, 5k̂.

2 marks: “Show that A(2, −1, 4), B(3, 1, 2) and C(5, 5, −2) are collinear.” AB⃗ = î + 2ĵ − 2k̂ and AC⃗ = 3î + 6ĵ − 6k̂ = 3AB⃗. The vectors are parallel and share A, so the points are collinear.

Remember it

Head minus tail gives the vector joining two points.