QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 202 Original QCE Vault practice · 1 mark
Browse all questions Vectors revision Original practice exam Report an issue Q202 · Practice question Technology-free Complex familiar 1 mark
QUESTION 202 Which plane contains all points equidistant from A = ( 1 , 0 , 2 ) A=(1,0,2) A = ( 1 , 0 , 2 ) and B = ( 3 , 2 , 0 ) B=(3,2,0) B = ( 3 , 2 , 0 ) ? WORKED SOLUTION
Answer C 1 mark Worked solution
The bisector plane passes through midpoint (2,1,1), with normal B-A=(2,2,-2). Its equation is 2 ( x − 2 ) + 2 ( y − 1 ) − 2 ( z − 1 ) = 0 2(x-2)+2(y-1)-2(z-1)=0 2 ( x − 2 ) + 2 ( y − 1 ) − 2 ( z − 1 ) = 0 , or x + y − z = 2 x+y-z=2 x + y − z = 2 . Equal squared distances give the same equation; D describes a line rather than the full plane. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus Related questions Q155 · Original practice · 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. Vectors in two and three dimensions Q188 · Original practice · 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 . Vectors in two and three dimensions Q189 · Original practice · 5 marks A triangular panel has vertices A = ( 1 , 0 , 0 ) A=(1,0,0) A = ( 1 , 0 , 0 ) , B = ( 0 , 2 , 0 ) B=(0,2,0) B = ( 0 , 2 , 0 ) and C = ( 0 , 0 , 3 ) C=(0,0,3) C = ( 0 , 0 , 3 ) . Vectors in two and three dimensions Q190 · Original practice · 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 . Vectors in two and three dimensions