QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 155 Original QCE Vault practice · 6 marks
Browse all questions Vectors revision Original practice exam Report an issue Q155 · Practice question Technology-free Complex familiar 6 marks
QUESTION 155 (6 marks) A point P P P has coordinates ( 4 , 1 , 3 ) (4,1,3) ( 4 , 1 , 3 ) . A plane Π \Pi Π has equation x + 2 y + 2 z = 3. x+2y+2z=3. x + 2 y + 2 z = 3. The point H H H is the perpendicular projection of P P P onto Π \Pi Π . The point P ′ P' P ′ is the reflection of P P P in Π \Pi Π , so H H H is the midpoint of P P ′ PP' P P ′ . The diagram is schematic. a) Determine the coordinates of H H H using a normal vector to Π \Pi Π . [3 marks] b) Determine the coordinates of P ′ P\prime P ′ and the length P P ′ PP\prime P P ′ . [3 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) H = ( 3 , − 1 , 1 ) H=(3,-1,1) H = ( 3 , − 1 , 1 ) . (b) P ′ = ( 2 , − 3 , − 1 ) P\prime=(2,-3,-1) P ′ = ( 2 , − 3 , − 1 ) and P P ′ = 6 PP\prime=6 P P ′ = 6 units. Worked solution
(a) A normal vector is n = ( 1 , 2 , 2 ) \mathbf n=(1,2,2) n = ( 1 , 2 , 2 ) , with n ⋅ n = 9 \mathbf n\cdot\mathbf n=9 n ⋅ n = 9 . Write h = p − λ n \mathbf h=\mathbf p-\lambda\mathbf n h = p − λ n . Because H H H is on the plane, ( 4 − λ ) + 2 ( 1 − 2 λ ) + 2 ( 3 − 2 λ ) = 3. (4-\lambda)+2(1-2\lambda)+2(3-2\lambda)=3. ( 4 − λ ) + 2 ( 1 − 2 λ ) + 2 ( 3 − 2 λ ) = 3. Thus 12 − 9 λ = 3 12-9\lambda=3 12 − 9 λ = 3 , so λ = 1 \lambda=1 λ = 1 and H = ( 3 , − 1 , 1 ) H=(3,-1,1) H = ( 3 , − 1 , 1 ) . (b) The midpoint relation gives
p ′ = 2 h − p = ( 2 , − 3 , − 1 ) . \mathbf p'=2\mathbf h-\mathbf p=(2,-3,-1). p ′ = 2 h − p = ( 2 , − 3 , − 1 ) . Also P H = ∣ n ∣ = 3 PH=|\mathbf n|=3 P H = ∣ n ∣ = 3 , so P P ′ = 2 P H = 6 PP'=2PH=6 P P ′ = 2 P H = 6 . As a check, p − h = h − p ′ = n \mathbf p-\mathbf h=\mathbf h-\mathbf p'=\mathbf n p − h = h − p ′ = n , proving perpendicularity and equal lengths. The reflected point need not lie on the plane. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Identify a normal vector (1,2,2).
[1 mark] Part a: Use the normal line and plane to obtain the parameter 1.
[1 mark] Part a: Obtain H=(3,-1,1).
[1 mark] Part b: Use the midpoint relation for the reflected point.
[1 mark] Part b: Obtain P-prime=(2,-3,-1).
[1 mark] Part b: Obtain PP-prime=6 units.
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