A-LevelMathematicsVectors9709/12 · Oct/Nov 2017 · Q99 Marks

The diagram shows a trapezium OABC in which OA is parallel to CB. The position vectors of A and B relative to the origin O are given by OA = (2, -2, -1) and OB = (6, 6, 1). [Figure X.X]

(i)Show that angle OAB is 90°.[3]
(ii)The magnitude of CB is three times the magnitude of OA. Find the position vector of C.[3]
(iii)Find the exact area of the trapezium OABC, giving your answer in the form a√b, where a and b are integers.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

AB = B - A or BA = A - B (M1) Use of b - a or a - b
e.g. AO . AB = -8 + 6 + 2 = 0 → OAB = 90° AG (M1A1) Use of dot product with either AO or OA & either AB or BA. Must see 3 component products.
OR OA = 3, OB = √38, AB = √29 (M1) Correct use of Pythagoras.
OA² + AB² = OB² → OAB = 90° AG (A1) In both methods must state angle or θ = 90° or similar for A1
CB = (-6, 6, -3) or BC = (6, -6, 3) (B1) Must correctly identify the vector.
OC = OB + BC (or -CB) = (0, 7, 4) (M1A1) Correct link leading to OC
OA = 3, BC = 9, AB = √29 (5.39) (B1) For any one of these
Area = ½(3 + 9)√29 or 3√29 + 3√29 (M1) Correct formula(e) used for trapezium or (rectangle + triangle) or two triangles using their lengths.
= 6√29 (1/1044, 2√261 or 3√116) (A1) Exact answer in correct form.

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