9 min readConcept sheet2-mark question; the first step of most 3 and 5-mark line questions

Vector equation of a line

The line through the point with position vector a⃗, parallel to the vector b⃗, is r⃗ = a⃗ + λb⃗. Each real value of λ gives one point.

Point and direction formr⃗ = a⃗ + λb⃗ (a⃗: position vector of a known point, b⃗: a vector along the line, λ: any real number)

Two point formr⃗ = a⃗ + λ(b⃗ − a⃗), where a⃗ and b⃗ are the position vectors of the two points

Direction ratiosIf b⃗ = aî + bĵ + ck̂, then a, b, c are direction ratios of the line

Testing a pointThe point lies on the line only if one value of λ fits all three components

Derivation: line through a point, parallel to a vector

Let the line l pass through A, with position vector a⃗, and be parallel to b⃗. Let P, with position vector r⃗, be any point on l.

1
AP⃗ = OP⃗ − OA⃗ = r⃗ − a⃗.
2
AP⃗ lies along l, so it is parallel to b⃗.
3
Parallel vectors are scalar multiples: AP⃗ = λb⃗ for some real λ.
4
So r⃗ − a⃗ = λb⃗.
5★
So, r⃗ = a⃗ + λb⃗ is the vector equation of the line.

★ key step

Line through two points

Let the line pass through A and B, with position vectors a⃗ and b⃗.

1
AB⃗ = b⃗ − a⃗ lies along the line.
2
Use the point A and the direction b⃗ − a⃗ in r⃗ = a⃗ + λ(direction).
3★
So, r⃗ = a⃗ + λ(b⃗ − a⃗). λ = 0 gives A and λ = 1 gives B.

★ key step

Example 1: point and direction

Problem

Find the vector equation of the line through (1, 2, −1) parallel to 3î − ĵ + 2k̂.

Working

a⃗ = î + 2ĵ − k̂ and b⃗ = 3î − ĵ + 2k̂.
Formula: r⃗ = a⃗ + λb⃗.
r⃗ = (î + 2ĵ − k̂) + λ(3î − ĵ + 2k̂)

Answer

r⃗ = (î + 2ĵ − k̂) + λ(3î − ĵ + 2k̂), where λ is a real number.

Example 2: through two points, and testing a point

Problem

Find the vector equation of the line through A(2, −1, 4) and B(1, 1, −2). Does it pass through C(0, 3, −8)?

Working

a⃗ = 2î − ĵ + 4k̂, b⃗ = î + ĵ − 2k̂.
b⃗ − a⃗ = (1 − 2)î + (1 + 1)ĵ + (−2 − 4)k̂ = −î + 2ĵ − 6k̂
r⃗ = (2î − ĵ + 4k̂) + λ(−î + 2ĵ − 6k̂)
A general point is (2 − λ, −1 + 2λ, 4 − 6λ).
For C: 2 − λ = 0 gives λ = 2.
Check: −1 + 2(2) = 3 and 4 − 6(2) = −8. Both match C.

Answer

r⃗ = (2î − ĵ + 4k̂) + λ(−î + 2ĵ − 6k̂). Yes, C lies on it (λ = 2).

Points to remember

The vector equation is not unique: any point on the line and any non-zero multiple of b⃗ give the same line.
Always write r⃗ = on the left; 'a⃗ + λb⃗' alone loses a mark.
λ is a real number, so write 'λ ∈ R' or 'λ is a scalar'.
To test a point, find λ from one component and check the other two.

In real life

A laser level used by masons on building sites shines a beam that follows r⃗ = a⃗ + λb⃗: a⃗ is where the device sits and b⃗ is the beam's direction.

Exam angle

1 marks: “Write a vector parallel to the line r⃗ = (2î − ĵ) + λ(î + 3k̂).” b⃗ = î + 3k̂, or any non-zero multiple of it.

2 marks: “Find the vector equation of the line through (3, −1, 2) parallel to 2î + 5ĵ − k̂.” a⃗ = 3î − ĵ + 2k̂, b⃗ = 2î + 5ĵ − k̂. So r⃗ = (3î − ĵ + 2k̂) + λ(2î + 5ĵ − k̂).

3 marks: “Find the vector equation of the line through A(1, −2, 3) and B(3, 0, 4). Does it pass through C(7, 4, 6)?” b⃗ − a⃗ = 2î + 2ĵ + k̂, so r⃗ = (î − 2ĵ + 3k̂) + λ(2î + 2ĵ + k̂). For C: 1 + 2λ = 7 gives λ = 3; then −2 + 6 = 4 and 3 + 3 = 6 match. Yes, C lies on it.

Remember it

Start + λ × direction: r⃗ = a⃗ + λb⃗.