Processing math: 100%

User interface language: English | Español

Date November 2019 Marks available 2 Reference code 19N.1.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_11 Adapted from N/A

Question

Points A(0 , 0 , 10) , B(0 , 10 , 0) , C(10 , 0 , 0) , V(p , p , p) form the vertices of a tetrahedron.

Consider the case where the faces ABV and ACV are perpendicular.

The following diagram shows the graph of θ against p. The maximum point is shown by X.

Show that AB×AV=10(102ppp) and find a similar expression for AC×AV.

[3]
a.i.

Hence, show that, if the angle between the faces ABV and ACV is θ, then cosθ=p(3p20)6p240p+100.

[5]
a.ii.

Find the two possible coordinates of V.

[3]
b.i.

Comment on the positions of V in relation to the plane ABC.

[1]
b.ii.

At X, find the value of p and the value of θ.

[3]
c.i.

Find the equation of the horizontal asymptote of the graph.

[2]
c.ii.

Markscheme

AV=(ppp10)      A1

AB×AV=(01010)×(ppp10)=(10(p10)+10p10p10p)      A1

=(20p10010p10p)=10(102ppp)      AG

AC×AV=(10010)×(ppp10)=(10p10020p10p)(=10(p102pp))      A1

 

[3 marks]

a.i.

attempt to find a scalar product        M1

10(102ppp)10(p102pp)=100(3p220p)

OR  (102ppp)(p102pp)=3p220p      A1

attempt to find magnitude of either AB×AV  or  AC×AV        M1

|10(102ppp)|=|10(p102pp)|=10(102p)2+2p2        A1

100(3p220p)=100((102p)2+2p2)2cosθ

cosθ=3p220p(102p)2+2p2        A1

Note: Award A1 for any intermediate step leading to the correct answer.

=p(3p20)6p240p+100      AG

Note: Do not allow FT marks from part (a)(i).

[8 marks]

a.ii.

p(3p20)=0p=0  or  p=203        M1A1

coordinates are (0, 0, 0) and (203203203)      A1

Note: Do not allow column vectors for the final A mark.

[3 marks]

b.i.

two points are mirror images in the plane
or opposite sides of the plane
or equidistant from the plane
or the line connecting the two Vs is perpendicular to the plane       R1

[1 mark]

b.ii.

geometrical consideration or attempt to solve 1=p(3p20)6p240p+100       (M1)

p=103θ=π  or  θ=180       A1A1

[3 marks]

c.i.

pcosθ12         M1

hence the asymptote has equation θ=π3        A1

[2 marks]

c.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.18—Intersections of lines & planes
Show 55 related questions
Topic 3— Geometry and trigonometry

View options