The vector OA has a magnitude of 15 units and is in the same direction as the vector 3i – 4k. The vector OB has a magnitude of 14 units and is in the same direction as the vector 2i + 3j – 6k.

Mark scheme

Marking points from the official mark scheme (May/June 2012). The mark or mark type is shown in brackets.

1. [M1] Dot product (3i – 4k) • (2i + 3j – 6k) = (3)(2) + (0)(3) + (-4)(-6) = 6 + 0 + 24 = 30; Uses x1x2 + y1y2 + z1z2.
2. [M1] Magnitudes: |3i – 4k| = √(3² + (-4)²) = √25 = 5. |2i + 3j – 6k| = √(2² + 3² + (-6)²) = √49 = 7; Method for modulus.
3. [M1 A1] cos θ = 30 / (5 × 7) = 30/35 = 6/7. → angle = 31.0° (1 d.p.) or 0.541 radians (3 s.f.); Links everything correctly. co.
4. [M1 A1] For OA: Find unit vector of (3i – 4k) then multiply by 15. OA = 15 * (3i – 4k)/5 = 9i – 12k; M mark for ×(15 ÷ 5) or ×(14 ÷ 7), A1 for OA.
5. [A1] For OB: Find unit vector of (2i + 3j – 6k) then multiply by 14. OB = 14 * (2i + 3j – 6k)/7 = 4i + 6j – 12k; A1 for OB.
6. [M1] Vector AB = OB - OA = (4i + 6j – 12k) - (9i – 12k) = -5i + 6j; Correct use for either AB or BA.
7. [M1] Magnitude of AB = √((-5)² + 6²) = √61 or 7.81; Complete method for unit vector.
8. [A1] Unit vector in direction of AB = (-5i + 6j) / √61; co.