A-LevelMathematicsVectors9709/12 · May/June 2017 · Q88 Marks
Relative to an origin O, the position vectors of three points A, B and C are given by OA = 3i + pj – 2pk, OB = 6i + (p + 4)j + 3k and OC = (p − 1)i + 2j + qk, where p and q are constants.
- (i)In the case where p = 2, use a scalar product to find angle AOB.[4]
- (ii)In the case where AB is parallel to OC, find the values of p and q.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- Uses scalar product correctly: 3×6+2×6+ (-4) × 3 = 18
- [M1] Use of dot product with OA or AO & OB or BO only.
- |OA| = √29, |OB| = 9
- [M1] Correct method for any one of |OA|, |AO|, |OB| or |BO|.
- √29 × 9 × cos AOB = 18
- [M1] All linked correctly.
- → AOB = 68.2° or 1.19 rad
- [A1] Multiples of π are acceptable (e.g. 0.379π°)
- AB = 3i + 4j +(3+2p)k
- Comparing "j"
- [DM1] For comparing, OC must contain p & q. Can be implied by AB = λOC.
- → p = 2½ and q = 4
- [A1 A1] Accuracy marks only available if AB is correct.
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