A-LevelFurther MathematicsVectors9231/12 · Oct/Nov 2020 · Q715 Marks

The points A, B, C have position vectors -2i+2j-k, -2i+j+2k, -2j+k, respectively, relative to the origin O.

(a)Find the equation of the plane ABC, giving your answer in the form ax+by+cz = d.[5]
(b)Find the acute angle between the planes OBC and ABC.[4]
(c)The point D has position vector ti-j. Given that the shortest distance between the lines AB and CD is √10, find the value of t.[6]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[B1] Finds direction vectors of two lines in the plane. AB = -j + 3k BC = 2i-3j-k
[M1 A1] Finds normal to the plane ABC. OE i j k; 0 -1 3; 2 -4 2 = 10i+6j+2k
[M1 A1] Substitutes point. OE 5(0) + 3(-2)+1(1) = -5 ⇒ 5x + 3y + z = -5
[M1 A1] Finds normal to the plane OBC. n_OBC = 5i + 2j + 4k
[M1] Uses dot product of normal vectors.
[A1] 0.49(0) rad or 28.1°
[M1 A1] Finds common perpendicular. AB = (0, -1, 3); CD = (t, 1, -1); n = AB x CD = (-2, 3t, t); AC = (2, -4, 2); |AC ⋅ n| / |n| = |-4 - 10t| / √(4 + 10t²)
[M1] Sets equal to √10 and solves for t. |-4-10t| / √(4+10t²) = √10
[A1] t = -4/3

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