Date | November 2018 | Marks available | 3 | Reference code | 18N.1.AHL.TZ0.H_9 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | H_9 | Adapted from | N/A |
Question
Consider a triangle OAB such that O has coordinates (0, 0, 0), A has coordinates (0, 1, 2) and B has coordinates (2b, 0, b − 1) where b < 0.
Let M be the midpoint of the line segment [OB].
Find, in terms of b, a Cartesian equation of the plane Π containing this triangle.
Find, in terms of b, the equation of the line L which passes through M and is perpendicular to the plane П.
Show that L does not intersect the y-axis for any negative value of b.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
n=(012)×(2b0b−1) (M1)
=(b−14b−2b) (M1)A1
(0, 0, 0) on Π so (b−1)x+4by−2bz=0 (M1)A1
METHOD 2
using equation of the form px+qy+rz=0 (M1)
(0, 1, 2) on Π ⇒ q+2r=0
(2b, 0, b − 1) on Π ⇒ 2bp+r(b−1)=0 (M1)A1
Note: Award (M1)A1 for both equations seen.
solve for p, q and r (M1)
(b−1)x+4by−2bz=0 A1
[5 marks]
M has coordinates (b,0,b−12) (A1)
r = (b0b−12)+λ(b−14b−2b) M1A1
Note: Award M1A0 if r = (or equivalent) is not seen.
Note: Allow equivalent forms such as x−bb−1=y4b=2z−b+1−4b.
[3 marks]
METHOD 1
x=z=0 (M1)
Note: Award M1 for either x=0 or z=0 or both.
b+λ(b−1)=0 and b−12−2λb=0 A1
attempt to eliminate λ M1
⇒−bb−1=b−14b (A1)
−4b2=(b−1)2 A1
EITHER
consideration of the signs of LHS and RHS (M1)
the LHS is negative and the RHS must be positive (or equivalent statement) R1
OR
−4b2=b2−2b+1
⇒5b2−2b+1=0
Δ=(−2)2−4×5×1=−16(<0) M1
∴ no real solutions R1
THEN
so no point of intersection AG
METHOD 2
x=z=0 (M1)
Note: Award M1 for either x=0 or z=0 or both.
b+λ(b−1)=0 and b−12−2λb=0 A1
attempt to eliminate b M1
⇒λ1+λ=11−4λ (A1)
−4λ2=1(⇒λ2=−14) A1
consideration of the signs of LHS and RHS (M1)
there are no real solutions (or equivalent statement) R1
so no point of intersection AG
[7 marks]