A-LevelMathematicsVectors9709/12 · Oct/Nov 2016 · Q910 Marks
Relative to an origin O, the position vectors of the points A, B and C are given by OA = (-2, 3, -1), OB = (-2, 6, 5) and OC = (2, -2, 6).
- (i)Use a scalar product to find angle AOB.[4]
- (ii)Find the vector which is in the same direction as AC and of magnitude 15 units.[3]
- (iii)Find the value of the constant p for which pOA + OC is perpendicular to OB.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.
- [M1] -4-6-6 = -16 Use of x1x2 + y1y2 + z1z2 on their OA & OB
- [M1] Modulus once on either their OA or OB
- [M1] All linked using their OA & OB
- [A1] → θ = 139.6° or 2.44° or 0.776π
- [B1] AC = c – a = (4, -5, 7) Magnitude = 10
- [M1] Scaling → 15/their10 × (0, 8, 6) = (0, 12, 9) For 15 × their unit vector.
- [A1]
- [B1] (2+2p, 6-2p, 5-p) Single vector soi by scalar product.
- [M1] Dot product of (p OA + OC) and OB = 0.
- [A1] → p=23/4
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