QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 188 Original QCE Vault practice · 5 marks
Browse all questions Vectors revision Original practice exam Report an issue Q188 · Practice question Technology-free Complex familiar 5 marks
QUESTION 188 (5 marks) A point P = ( 2 , 1 , 2 ) P=(2,1,2) P = ( 2 , 1 , 2 ) and a line L : r = ( 1 , 0 , 0 ) + t ( 1 , 2 , 2 ) L:\mathbf r=(1,0,0)+t(1,2,2) L : r = ( 1 , 0 , 0 ) + t ( 1 , 2 , 2 ) are given, with t ∈ R t\in\mathbb R t ∈ R . a) Determine the point H on L nearest to P.
[3 marks] b) Determine the exact distance PH and verify perpendicularity.
[2 marks] WORKED SOLUTION
Practice marking scheme 5 marks ANSWER (a) H = ( 16 / 9 , 14 / 9 , 14 / 9 ) H=(16/9,14/9,14/9) H = ( 16/9 , 14/9 , 14/9 ) . (b) 5 / 3 \sqrt5/3 5 /3 . Worked solution
(a) Let a = ( 1 , 0 , 0 ) \mathbf a=(1,0,0) a = ( 1 , 0 , 0 ) and d = ( 1 , 2 , 2 ) \mathbf d=(1,2,2) d = ( 1 , 2 , 2 ) . Perpendicularity requires ( p − a − t d ) ⋅ d = 0 (\mathbf p-\mathbf a-t\mathbf d)\cdot\mathbf d=0 ( p − a − t d ) ⋅ d = 0 , so t = 7 / 9 t=7/9 t = 7/9 . Substitution gives H. The squared-distance quadratic has positive leading coefficient, so this is its unique minimum. (b) p − h = ( 2 , − 5 , 4 ) / 9 \mathbf p-\mathbf h=(2,-5,4)/9 p − h = ( 2 , − 5 , 4 ) /9 has magnitude 45 / 9 = 5 / 3 \sqrt{45}/9=\sqrt5/3 45 /9 = 5 /3 . Its dot product with ( 1 , 2 , 2 ) (1,2,2) ( 1 , 2 , 2 ) is ( 2 − 10 + 8 ) / 9 = 0 (2-10+8)/9=0 ( 2 − 10 + 8 ) /9 = 0 . Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Form the perpendicular projection condition.
[1 mark] Part a: Give all coordinates of H.
[1 mark] Part b: Calculate the distance.
[1 mark] Part b: Check the zero dot product.
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