The line l has equation r = i – 2j – 3k + λ(−i + j + 2k). The points A and B have position vectors –2i + 2j – k and 3i – j + k respectively.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [M1] (a) Use correct process for modulus on direction vector of l, e.g. √((-1)² +1² +2²).
2. [A1] (a) Obtain (1/√6)(-i+j+2k). OE. Allow coordinates as row or column, but not row or column with i, j and k included.
3. [M1] (b) Use a correct method to form an equation for line m. Allow even if all signs of point incorrect, namely use +2i - 2j + k or -3i + j – k.
4. [A1] (b) Obtain r = -2i + 2j – k + μ(−5i + 3j – 2k). OE, e.g. r = 3i – j + k + μ(−5i + 3j − 2k). Must have r = ...
5. [B1] (c) Justify lines are not parallel. (-5, 3, -2) ≠ d (− 1, 1, 2) or ( −5, 3, -2)x(-1, 1, 2) ≠ 0. Can find angle (105°, 74.6°, 1.84º or 1.3(0)º) instead but if incorrect B0 and A0 at end. Accept direction vectors don't have common factor but not direction vectors are not equal or direction vectors are different or μ ≠ λ or scalar product ≠ 0. Not the line equations are not multiples of each other.
6. [B1] (c) Express l or m in component form. e.g. (1 – λ, -2 + λ, −3 + 2λ) and (-2 – 5μ, 2 + 3μ, −1 – 2μ).
7. [M1] (c) Equate two pairs of components of general points on l and their m and solve simultaneously for λ or for μ.
8. [A1] (c) Obtain correct answer for λ or μ, e.g. λ = 1/2, μ = 1/2.
9. [A1] (c) Determine that all three equations are not satisfied and the lines fail to intersect and conclude the lines are skew. Conclusion needs to follow correct working.