A-LevelMathematicsVectors9709/12 · May/June 2019 · Q88 Marks

The position vectors of points A and B, relative to an origin O, are given by OA = (6, -2, -6) and OB = (3, k, -3), where k is a constant.

(i)Find the value of k for which angle AOB is 90°.[2]
(ii)Find the values of k for which the lengths of OA and OB are equal.[2]
(iii)The point C is such that AC = 2CB. In the case where k = 4, find the unit vector in the direction of OC.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[M1] 6×3+-2xk+-6×-3 = 0 (18-2k+18= 0)
[A1] k = 18
[M1] 76 + 18 + k² = 18 + (k + 2)²
[A1] k = 18
[M1] 36+4+36= 9+k²+9
[A1] k = ±√58 or ±7.62
[M1] Complete method using AC = ± 2/3 AB And then OA + their AC
[A1] OC = (4, 2, -4)
[M1] Divides by modulus of their OC
[A1] 1/6 (4i + 2j – 4k) or 1/6 (4,2,-4)
[M1] Let OC = (p,q,r) → AC = (p-6, q+2, r+6) & CB = (3-p, k-q, -3-r). Equates coefficients leading to values for p, q, r
[A1] →p=4, q=2 & r= −4
[M1] Divides by modulus of their OC
[A1] 1/6 (4i + 2j – 4k) or 1/6 (4,2,-4)
[M1] CB = OB – OC :. 2(OB-OC) = OC – OA → 2OB + OA = 3OC
[A1] 3OC = (12, 6, -12) → OC = (4, 2, -4)
[M1] Divides by modulus of their OC
[A1] 1/6 (4i + 2j – 4k) or 1/6 (4,2,-4)

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