Relative to an origin O, points A and B have position vectors (5 -6) and (29 -13) respectively.
- (i)Find a unit vector parallel to AB.[3]
- (ii)The points A, B and C lie on a straight line such that 2AC = 3AB. Find the position vector of the point C.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- (i) (29 - 5)/(-13 - (-6)) = (24 -7); M1 for subtraction
- Magnitude = 25, unit vector (1/25) (24 -7); M1 for attempt to find magnitude of their vector; A1
- (ii) 2AC = 3AB; M1 for attempt to find AC – may be part of a larger method
- or 2AB+2BC = 3AB leading to AC = (36 -10.5)
- OC = OA + AC; M1 for attempt to find OC
- or OB-OA = 2OC – 2OB leading to OC = (41 -16.5)
- A1 for each; A1
- (equivalent methods acceptable)
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