Date | November 2020 | Marks available | 4 | Reference code | 20N.2.AHL.TZ0.H_10 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | H_10 | Adapted from | N/A |
Question
The plane Π1Π1 has equation 3x−y+z=−133x−y+z=−13 and the line LL has vector equation
r=(12-2)+λ(-3-14) , λ ∈ ℝ.
The plane Π2 contains the point O and the line L.
Given that L meets Π1 at the point P, find the coordinates of P.
Find the shortest distance from the point O(0, 0, 0) to Π1.
Find the equation of Π2, giving your answer in the form r.n=d.
Determine the acute angle between Π1 and Π2.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3(1-3λ)-(2-λ)+(-2+4λ)=-13 (M1)
λ=3 (A1)
r=(12-2)+3(-3-14)=(8-110) (M1)
so P(-8, -1, 10) A1
Note: Do not award the final A1 if a vector given instead of coordinates
[4 marks]
METHOD 1
r=μ(3-11)
substituting into equation of the plane M1
9μ+μ+μ=-13
μ=-1311 (=-1.18…) A1
distance =13√32+(-1)2+1211 (M1)
=13√11(=13√1111=3.92) A1
METHOD 2
choice of any point on the plane, eg (-8, -1, 10) to use in distance formula (M1)
so distance =(-8-110)·(-31-1)√(-3)2+12+(-1)2 A1A1
Note: Award A1 for numerator, A1 for denominator.
=24-1-10√11
=13√11(=13√1111=3.92) A1
[4 marks]
EITHER
identify two vectors (A1)
eg, (12-2) and (-3-14)
n=(12-2)×(-3-14)=(625) (M1)
OR
identify three points in the plane (A1)
eg λ=0,1 gives (12-2) and (-212)
solving system of equations (M1)
THEN
Π2 : r.(625)=0 A1
Note: Accept 6x+2y+5z=0.
[3 marks]
vector normal to Π1 is eg n1=(3-11)
vector normal to Π2 is eg n2=(625) (A1)
required angle is θ, where cos θ(3-11)·(625)√11√65 M1A1
cos θ=21√11√65=0.785… (A1)
θ=0.667526…
θ=0.668 (=38.2°) A1
Note: Award the penultimate (A1) but not the final A1 for the obtuse angle 2.47406… or 142°.
[5 marks]