QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 191 Original QCE Vault practice · 6 marks
Browse all questions Vectors revision Original practice exam Report an issue Q191 · Practice question Technology-free Complex unfamiliar 6 marks
QUESTION 191 (6 marks) Two straight rails in a sculpture are modelled by L 1 : ( x , y , z ) = ( t , t , 0 ) L_1:(x,y,z)=(t,t,0) L 1 : ( x , y , z ) = ( t , t , 0 ) and L 2 : ( x , y , z ) = ( s , 1 − s , 1 ) L_2:(x,y,z)=(s,1-s,1) L 2 : ( x , y , z ) = ( s , 1 − s , 1 ) , with s , t ∈ R s,t\in\mathbb R s , t ∈ R . a) Show that the lines are neither parallel nor intersecting.
[2 marks] b) Determine their minimum separation and both ends of the shortest connector.
[4 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) The lines are skew. (b) Distance 1 1 1 ; endpoints ( 1 / 2 , 1 / 2 , 0 ) , ( 1 / 2 , 1 / 2 , 1 ) (1/2,1/2,0),(1/2,1/2,1) ( 1/2 , 1/2 , 0 ) , ( 1/2 , 1/2 , 1 ) . Worked solution
(a) Directions ( 1 , 1 , 0 ) (1,1,0) ( 1 , 1 , 0 ) and ( 1 , − 1 , 0 ) (1,-1,0) ( 1 , − 1 , 0 ) are not scalar multiples. The lines have different constant z coordinates, so they cannot intersect. (b) For arbitrary points, d 2 = ( t − s ) 2 + ( t + s − 1 ) 2 + 1 ≥ 1 d^2=(t-s)^2+(t+s-1)^2+1\ge1 d 2 = ( t − s ) 2 + ( t + s − 1 ) 2 + 1 ≥ 1 . Equality requires t=s and t+s=1, giving t=s=1/2. The vertical connector between the stated points has length 1 and is perpendicular to both line directions. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Rule out parallel directions.
[1 mark] Part a: Rule out intersection.
[1 mark] Part b: Form distance with independent parameters.
[1 mark] Part b: Establish the global lower bound.
[1 mark] Part b: Solve both equality conditions.
[1 mark] Part b: Give both endpoints and the distance.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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