A-Level Maths vectors: 3D lines, intersections and the scalar product

Where vectors sit in each specification

The key ideas

A position vector gives a point relative to the origin O. The vector from A to B is AB = b - a. The magnitude of (x, y, z) is sqrt(x^2 + y^2 + z^2), and a unit vector is the vector divided by its magnitude.

A line through a point with position vector a in direction b is r = a + tb, where t is a parameter. To test two lines, first compare directions: if one direction is a multiple of the other, the lines are parallel. If not, set the general points equal, solve two of the three component equations for the two parameters, and check the third. If it works, the lines intersect at that point; if not, they are skew.

The scalar product a.b = a1b1 + a2b2 + a3b3 also equals |a||b| cos θ, so cos θ = a.b/(|a||b|). For the angle between two lines use their direction vectors. If a.b = 0 the vectors are perpendicular, which is how you find the foot of the perpendicular from a point to a line: the vector from the point to a general point on the line must be perpendicular to the line's direction.

Worked example 1: the line through A(1, 2, -1) and B(3, 1, 2), and its angle with the direction (1, 1, 1)

1. Direction: AB = b - a = (3 - 1, 1 - 2, 2 - (-1)) = (2, -1, 3).
2. Equation: r = (1, 2, -1) + t(2, -1, 3).
3. Scalar product with (1, 1, 1): 2 - 1 + 3 = 4.
4. Magnitudes: |(2, -1, 3)| = sqrt14 and |(1, 1, 1)| = sqrt3.
5. cos θ = 4/sqrt42 = 0.6172, so θ = 51.9°.

Worked example 2: do the lines r = (1, 2, -1) + t(2, -1, 3) and r = (1, 3, 4) + s(1, -1, -1) intersect?

1. The directions (2, -1, 3) and (1, -1, -1) are not multiples of each other, so the lines are not parallel.
2. Equate components: 1 + 2t = 1 + s, 2 - t = 3 - s, -1 + 3t = 4 - s.
3. From the first, s = 2t. Substitute in the second: 2 - t = 3 - 2t, so t = 1 and s = 2.
4. Check the third: left side -1 + 3 = 2, right side 4 - 2 = 2. They agree.
5. So the lines intersect, at t = 1: the point (3, 1, 2).

Worked example 3: the foot of the perpendicular from P(4, 3, 2) to the line r = (1, 2, -1) + t(2, -1, 3)

1. A general point on the line is Q = (1 + 2t, 2 - t, -1 + 3t).
2. PQ = Q - P = (2t - 3, -t - 1, 3t - 3).
3. For a perpendicular, PQ.(2, -1, 3) = 0: 2(2t - 3) - (-t - 1) + 3(3t - 3) = 4t - 6 + t + 1 + 9t - 9 = 14t - 14 = 0.
4. So t = 1 and the foot is (3, 1, 2).
5. The perpendicular distance is |PQ| = |(-1, -2, 0)| = sqrt5.

Common mistakes that cost marks

• Using position vectors instead of direction vectors when finding the angle between lines.
• Using the same parameter letter for both lines, so they can never be solved properly.
• Solving two equations and forgetting to check the third, then wrongly claiming the lines intersect.
• Calling non-parallel lines that do not meet 'parallel' instead of 'skew'.
• Sign errors in b - a, especially with negative components.
• Giving the acute angle when the obtuse is asked for, or the reverse. If cos θ is negative, the acute angle is 180° minus θ.

Exam technique and how a tutor helps

Vector questions are long and structured, often 8 to 10 marks across several parts that build on each other. Write vectors in column form or as (x, y, z) consistently, label every vector, and state your conclusion in words: 'the lines intersect at (3, 1, 2)' or 'the third equation is not satisfied, so the lines are skew'. A sketch, even a rough one, helps you see whether an answer is reasonable.

Students usually understand vectors but lose marks in the arithmetic. In one-to-one lessons a tutor works through the full multi-part questions with the student, checking each component on the shared whiteboard, and builds the habit of substituting answers back into all three equations before moving on.

Self-check: can you do these?

• Find |(2, -3, 6)|. (Answer: 7)
• Find a unit vector in the direction of (2, -3, 6). (Answer: (2/7, -3/7, 6/7))
• Find (1, 2, 3).(4, -5, 6). (Answer: 12)
• Are (1, 2, 0) and (2, -1, 3) perpendicular? (Answer: yes, the scalar product is 0)
• Write the equation of the line through (0, 1, 2) parallel to (1, 0, -1). (Answer: r = (0, 1, 2) + t(1, 0, -1))