Date | November Example questions | Marks available | 3 | Reference code | EXN.2.AHL.TZ0.11 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Show that | Question number | 11 | Adapted from | N/A |
Question
The points A(5, -2, 5), B(5, 4, -1), C(-1, -2, -1) and D(7, -4, -3) are the vertices of a right-pyramid.
The line L passes through the point D and is perpendicular to Π.
Find the vectors →AB and →AC.
Use a vector method to show that BˆAC=60°.
Show that the Cartesian equation of the plane Π that contains the triangle ABC is -x+y+z=-2.
Find a vector equation of the line L.
Hence determine the minimum distance, dmin, from D to Π.
Find the volume of right-pyramid ABCD.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
→AB=(06-6) (=6(01-1)) A1
→AC=(-60-6) (=6(-10-1)) A1
[2 marks]
attempts to use cos BˆAC=→AB·→AC|→AB||→AC| (M1)
=(06-6)·(-60-6)√72×√72 A1
=12 A1
so BˆAC=60° AG
[3 marks]
attempts to find a vector normal to Π M1
for example, →AB×→AC=(-363636) (=36(-111)) leading to A1
a vector normal to Π is n=(-111)
EITHER
substitutes (5, -2, -5) (or (5, 4, -1) or (-1, -2, -1)) into -x+y+z=d and attempts to find the value of d
for example, d=-5-2+5 (=-2) M1
OR
attempts to use r·n=a·n M1
for example, (xyz)·(-111)=(5-25)·(-111)
THEN
leading to the Cartesian equation of Π as -x+y+z=-2 AG
[3 marks]
r=(7-4-3)+λ(-111) (λ∈ℝ) A1
[1 mark]
substitutes x=7-λ, y=-4+λ, z=-3+λ into -x+y+z=-2 (M1)
-(7-λ)+(-4+λ)+(-3+λ)=-2 (3λ=12)
λ=4 A1
shows a correct calculation for finding dmin, for example, attempts to find
|4(-111)| M1
dmin=4√3 (=6.93) A1
[4 marks]
let the area of triangle ABC be A
EITHER
attempts to find A=12|→AB×→AC|, for example M1
A=12|(-363636)|
OR
attempts to find 12|→AB||→AC|sin θ, for example M1
A=12×6√2×6√2×√32 (where sinπ3=√32)
THEN
A=18√3 (=31.2) A1
uses V=13Ah where A is the area of triangle ABC and h=dmin M1
V=13×18√3×4√3
=72 A1
[4 marks]