With respect to the origin O, the points A and B have position vectors given by OA = 2i – j and OB = j – 2k.
- (a)Show that OA = OB and use a scalar product to calculate angle AOB in degrees.[4]
- (b)Find the possible position vectors of P.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] Show that OA = OB = √5
- [M1] Evaluate the scalar product of the correct position vectors
- [M1] Divide their scalar product by the product of the moduli of their vectors and evaluate the inverse cosine of the result
- [A1] Obtain answer 101.5°
- [B1] State or imply M has position vector i – k
- [M1] Taking a general point of OM to have position vector λi – λk, express AP = √7 OA as an equation in λ
- [A1] State a correct equation in any form
- [A1] Reduce to λ² – 2λ − 15 = 0
- [DM1] Solve a quadratic and state a position vector
- [A1] Obtain answers 5i – 5k and −3i + 3k
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