QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 190 Original QCE Vault practice · 8 marks
Browse all questions Vectors revision Original practice exam Report an issue Q190 · Practice question Technology-active Complex unfamiliar 8 marks
QUESTION 190 (8 marks) A laser ray follows r = ( 1 , 0 , 2 ) + t ( 2 , 1 , − 1 ) \mathbf r=(1,0,2)+t(2,1,-1) r = ( 1 , 0 , 2 ) + t ( 2 , 1 , − 1 ) , t ≥ 0 t\ge0 t ≥ 0 . A reflecting screen lies in x + y + z = 6 x+y+z=6 x + y + z = 6 . a) Determine where the ray strikes the screen.
[3 marks] b) Determine the acute ray-screen angle to two decimal places in degrees.
[2 marks] c) After reflection, the direction is d o u t = d − 2 d ⋅ n n ⋅ n n \mathbf d_{\rm out}=\mathbf d-2\frac{\mathbf d\cdot\mathbf n}{\mathbf n\cdot\mathbf n}\mathbf n d out = d − 2 n ⋅ n d ⋅ n n , where d = ( 2 , 1 , − 1 ) \mathbf d=(2,1,-1) d = ( 2 , 1 , − 1 ) and n = ( 1 , 1 , 1 ) \mathbf n=(1,1,1) n = ( 1 , 1 , 1 ) . Find where the reflected ray first reaches z = − 3 z=-3 z = − 3 . [3 marks] WORKED SOLUTION
Practice marking scheme 8 marks ANSWER (a) ( 4 , 3 / 2 , 1 / 2 ) (4,3/2,1/2) ( 4 , 3/2 , 1/2 ) at t = 3 / 2 t=3/2 t = 3/2 . (b) 28.13 ∘ 28.13^\circ 28.1 3 ∘ . (c) ( 5 , 1 , − 3 ) (5,1,-3) ( 5 , 1 , − 3 ) . Worked solution
(a) Substitution gives 1 + 2 t + t + 2 − t = 6 1+2t+t+2-t=6 1 + 2 t + t + 2 − t = 6 , hence t=3/2. This is nonnegative, so the intersection lies on the ray. Substitution into the position gives the stated point. (b) With direction ( 2 , 1 , − 1 ) (2,1,-1) ( 2 , 1 , − 1 ) and normal ( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) , sin α = 2 / 18 = 2 / 3 \sin\alpha=2/\sqrt{18}=\sqrt2/3 sin α = 2/ 18 = 2 /3 . Thus α = 28.13 ∘ \alpha=28.13^\circ α = 28.1 3 ∘ . The angle to the normal is the complement of the requested angle. (c) The formula gives d o u t = ( 2 , 1 , − 1 ) − 4 3 ( 1 , 1 , 1 ) = ( 2 / 3 , − 1 / 3 , − 7 / 3 ) \mathbf d_{\rm out}=(2,1,-1)-\frac43(1,1,1)=(2/3,-1/3,-7/3) d out = ( 2 , 1 , − 1 ) − 3 4 ( 1 , 1 , 1 ) = ( 2/3 , − 1/3 , − 7/3 ) . The reflected ray starts at ( 4 , 3 / 2 , 1 / 2 ) (4,3/2,1/2) ( 4 , 3/2 , 1/2 ) and has parameter s ≥ 0 s\ge0 s ≥ 0 . Set 1 / 2 − 7 s / 3 = − 3 1/2-7s/3=-3 1/2 − 7 s /3 = − 3 , giving s = 3 / 2 s=3/2 s = 3/2 . Then x = 5 , y = 1 , z = − 3 x=5,y=1,z=-3 x = 5 , y = 1 , z = − 3 . Because the z component is strictly negative, this is the unique forward intersection. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Substitute into the plane.
[1 mark] Part a: Find and validate t=3/2.
[1 mark] Part a: Give the intersection point.
[1 mark] Part b: Use the line-plane scalar-product relation.
[1 mark] Part b: Calculate the acute angle.
[1 mark] Part c: Calculate the reflected direction.
[1 mark] Part c: Use the correct starting point and nonnegative parameter.
[1 mark] Part c: Find and validate all intersection coordinates.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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