To find an intersection, write each line with its own parameter, set the two position vectors equal, then compare coefficients of a and b to solve the two parameters. Substituting back confirms the point.
This lesson needs displacements and dividing points.
How do you solve an intersection step by step?
- Write each line as a position vector plus a parameter times a direction: for example a + λ(…).
- Expand so each expression looks like (…)a + (…)b.
- Equate the coefficients of a and of b. You now have two equations.
- Solve for λ and μ, then substitute one back for the point.
- Check with the other parameter.
Worked example
→OA = a, →OB = b. M is the midpoint of OA and N is on OB with ON = 2/3 b.
Lines AN and BM meet at X. Find →OX.
Line AN: →OX = a + λ(2/3 b − a) = (1 − λ)a + (2λ/3)b.
Line BM: →OM = ½a, so →OX = b + μ(½a − b) = (μ/2)a + (1 − μ)b.
Equate coefficients:
- a: 1 − λ = μ/2
- b: 2λ/3 = 1 − μ
From the first, μ = 2 − 2λ. Into the second: 2λ/3 = 1 − 2 + 2λ = −1 + 2λ. So −4λ/3 = −1, giving λ = 3/4 and μ = 1/2.
Answer: →OX = (1 − 3/4)a + (2 × 3/4 ÷ 3)b = ¼a + ½b.
Check with μ = 1/2: (1/4)a + (1 − 1/2)b = ¼a + ½b. Same, so X is correct.
The mistake to watch for
A common slip is to use one parameter for both lines.
Mistaken working: Line 1 gives 1 − λ = λ/2, so λ = 2/3. Then 2λ/3 = 1 − λ gives λ = 3/5. “Contradiction, so the lines do not meet.”
The student used λ in both lines.
The lines do meet, at different stages along each line. Two lines need two letters, λ and μ. When a solution seems impossible, first check that each line has its own parameter.
Check yourself
1. Line 1: r = (1, 0) + λ(1, 2). Line 2: r = (7, 0) + μ(−1, 1). Find the intersection.
Show answer
x: 1 + λ = 7 − μ. y: 2λ = μ. Substitute: 1 + λ = 7 − 2λ, so λ = 2, μ = 4. Check line 2: (7 − 4, 4) = (3, 4). Line 1: (1 + 2, 4) = (3, 4).
(3, 4)
2. Show that r = (2, 1) + s(1, 1) and r = (0, 3) + t(1, 1) do not meet.
Show answer
The directions are equal, so the lines are parallel. Is (2, 1) on the second line? Need 2 = t and 1 = 3 + t, so t = 2 and t = −2. Inconsistent.
Parallel and distinct, so no intersection.
3. →OA = a, →OB = b. P is the midpoint of AB and Q is on OB with OQ : QB = 1 : 2. Lines OP and AQ meet at X. Find →OX.
Show answer
OP: →OX = λ(½a + ½b). AQ: →OX = (1 − μ)a + (μ/3)b.
a: λ/2 = 1 − μ. b: λ/2 = μ/3. So 1 − μ = μ/3, μ = 3/4, λ = 1/2.
→OX = ¼a + ¼b
Where this leads next
The remaining piece is reading direction from an equation, in explain direction in a vector equation. Then test the whole module with the practice set.
If simultaneous-equation slips are costing you marks here, see how we approach this in online one-to-one Additional Mathematics tuition.