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Date May 2018 Marks available 5 Reference code 18M.1.AHL.TZ1.H_10
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Find Question number H_10 Adapted from N/A

Question

The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).

The Cartesian equation of the plane Π2Π2, passing through the points B , C and D , is y+z=1y+z=1.

The plane Π3Π3 passes through O and is normal to the line BD.

Π3Π3 cuts AD and BD at the points P and Q respectively.

Find the Cartesian equation of the plane Π1Π1, passing through the points A , B and D.

[3]
a.

Find the angle between the faces ABD and BCD.

[4]
b.

Find the Cartesian equation of Π3Π3.

[3]
c.

Show that P is the midpoint of AD.

[4]
d.

Find the area of the triangle OPQ.

[5]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognising normal to plane or attempting to find cross product of two vectors lying in the plane      (M1)

for example, AB×AD=(010)×(101)=(101)AB×AD=010×101=101     (A1)

Π1:x+z=1Π1:x+z=1     A1

[3 marks]

a.

EITHER

(101)(011)=1=22cosθ101011=1=22cosθ     M1A1

OR

|(101)×(011)|=3=22sinθ∣ ∣101×011∣ ∣=3=22sinθ     M1A1

Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.

θ=60(=π3)θ=60(=π3)     A1

angle between faces is 20(=2π3)20(=2π3)     A1

[4 marks]

b.

DB=(111)DB=111 or BD=(111)BD=111     (A1)

Π3:x+yz=kΠ3:x+yz=k     (M1)

Π3:x+yz=0Π3:x+yz=0     A1

[3 marks]

c.

METHOD 1

line AD : (r =)(001)+λ(101)001+λ101     M1A1

intersects Π3Π3 when λ(1λ)=0λ(1λ)=0     M1

so λ=12λ=12     A1

hence P is the midpoint of AD      AG

 

METHOD 2

midpoint of AD is (0.5, 0, 0.5)      (M1)A1

substitute into x+yz=0x+yz=0     M1

0.5 + 0.5 − 0.5 = 0     A1

hence P is the midpoint of AD     AG

[4 marks]

d.

METHOD 1

OP=12,OPQ=90,OQP=60OP=12,OPQ=90,OQP=60      A1A1A1

PQ=16PQ=16     A1

area =1212=143=312=1212=143=312     A1

 

METHOD 2

line BD : ( =)(110)+λ(111)110+λ111

λ=23λ=23     (A1)

OQ=(131323)OQ=⎜ ⎜ ⎜ ⎜ ⎜ ⎜131323⎟ ⎟ ⎟ ⎟ ⎟ ⎟    A1

area = 12|OP×OQ|12OP×OQ     M1

OP=(12012)OP=⎜ ⎜ ⎜ ⎜12012⎟ ⎟ ⎟ ⎟    A1

Note: This A1 is dependent on M1.

area = 312312     A1

[5 marks]

e.

Examiners report

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Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.17—Vector equations of a plane
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Topic 3— Geometry and trigonometry » AHL 3.18—Intersections of lines & planes
Topic 3— Geometry and trigonometry

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