Relative to an origin O, the position vectors of points A and B are given by OA = i + 2j and OB = 4i + pk.
- (i)In the case where p = 6, find the unit vector in the direction of AB.[3]
- (ii)Find the values of p for which angle AOB = cos⁻¹ (1/3).[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2013). The mark or mark type is shown in brackets.
- [1] AB = b − a = 3i − 2j + 6k
- [1] Divides by modulus.
- [1] Unit vector = (3i – 2j + 6k) / 7
- [1] Scalar product OA.OB = 4
- [1] Moduli calculated: |OA|=√5, |OB|=√(16+p²)
- [1] All linked correctly including correct use of cosθ=1/5 (e.g. 4 = √5 × √(16 + p²) × (1/5))
- [1] p = ±8
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