The line l₁ passes through the point (3, 1, −6) and is parallel to the vector 2i+j+4k. The line l₂ passes through the point (−1, 3, −6) and is perpendicular to the vector 3i-2j+k. The direction vector for l₂ has no component in the x-direction.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. Obtain r = 3i + j − 6k + λ (2i + j + 4k)
2. Carry out a correct method for finding a direction vector for l₂, (3i - 2j +k)·(yj + zk) = 0
3. Obtain r = -i + 3j – 6k + μ(j + 2k)
4. Carry out correct process for evaluating the scalar product of the direction vectors of l₁ and l₂
5. Using the correct process for the moduli, divide the scalar product by the product of the moduli and state cos θ = the result
6. Obtain answer 28.6° or 0.498°
7. Equate components of general points on their l₁ and their l₂, provided these are both equations of lines and solve for λ or for μ
8. Obtain correct answer for λ or μ, e.g. λ = −2, μ = −4
9. Obtain position vector of point of intersection is –i – j – 14k