QCE Vault / Specialist Maths Vectors in three dimensions — Question 26 Original QCE Vault practice · 6 marks
Browse all questions Vectors revision Original practice exam Report an issue Q26 · Practice question Complex familiar 6 marks
QUESTION 26 (6 marks) The sphere x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9 is intersected by the line r = ( 0 , 0 , 4 ) + t ( 1 , 0 , − 1 ) . \mathbf r=(0,0,4)+t(1,0,-1). r = ( 0 , 0 , 4 ) + t ( 1 , 0 , − 1 ) . a) Determine the two points where the line intersects the sphere.
[4 marks] b) Determine the exact distance between the two intersection points.
[2 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) ( 2 ± 2 / 2 , 0 , 2 ∓ 2 / 2 ) (2\pm\sqrt2/2,0,2\mp\sqrt2/2) ( 2 ± 2 /2 , 0 , 2 ∓ 2 /2 ) ; (b) 2 2 2 . Worked solution
(a) On the line, ( x , y , z ) = ( t , 0 , 4 − t ) (x,y,z)=(t,0,4-t) ( x , y , z ) = ( t , 0 , 4 − t ) . Substitution gives t 2 + ( 4 − t ) 2 = 9 ⇒ 2 t 2 − 8 t + 7 = 0. t^2+(4-t)^2=9\quad\Rightarrow\quad2t^2-8t+7=0. t 2 + ( 4 − t ) 2 = 9 ⇒ 2 t 2 − 8 t + 7 = 0. The quadratic formula gives t = 2 ± 2 / 2 t=2\pm\sqrt2/2 t = 2 ± 2 /2 . The points are ( 2 + 2 2 , 0 , 2 − 2 2 ) , ( 2 − 2 2 , 0 , 2 + 2 2 ) . \left(2+\frac{\sqrt2}{2},0,2-\frac{\sqrt2}{2}\right),\quad\left(2-\frac{\sqrt2}{2},0,2+\frac{\sqrt2}{2}\right). ( 2 + 2 2 , 0 , 2 − 2 2 ) , ( 2 − 2 2 , 0 , 2 + 2 2 ) . (b) Their displacement is ( 2 , 0 , − 2 ) (\sqrt2,0,-\sqrt2) ( 2 , 0 , − 2 ) , so the distance is 2 + 2 = 2 \sqrt{2+2}=2 2 + 2 = 2 . Substitutes the line into the sphere.
[1 mark] Solves the quadratic for t t t . [2 marks] Determines the exact distance.
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