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Date May 2018 Marks available 1 Reference code 18M.1.AHL.TZ2.H_9
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Explain Question number H_9 Adapted from N/A

Question

The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.

It is given that  AB = DC .

The position vectors  OA OB OC and  OD are given by

a = i + 2j − 3k

b = 3ij + pk

c = qi + j + 2k

d = −i + rj − 2k

where p , q and r are constants.

The point where the diagonals of ABCD intersect is denoted by M.

The plane Π cuts the x, y and z axes at X , Y and Z respectively.

Explain why ABCD is a parallelogram.

[1]
a.i.

Using vector algebra, show that AD = BC .

[3]
a.ii.

Show that p = 1, q = 1 and r = 4.

[5]
b.

Find the area of the parallelogram ABCD.

[4]
c.

Find the vector equation of the straight line passing through M and normal to the plane Π  containing ABCD.

[4]
d.

Find the Cartesian equation of Π .

[3]
e.

Find the coordinates of X, Y and Z.

[2]
f.i.

Find YZ.

[2]
f.ii.

Markscheme

a pair of opposite sides have equal length and are parallel      R1

hence ABCD is a parallelogram      AG

[1 mark]

a.i.

attempt to rewrite the given information in vector form       M1

ba = cd      A1

rearranging d − a = c − b       M1

hence   AD = BC      AG

Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.

[3 marks]

a.ii.

EITHER

use of  AB = DC      (M1)

( 2 3 p + 3 ) = ( q + 1 1 r 4 )        A1A1

OR

use of  AD = BC       (M1)

( 2 r 2 1 ) = ( q 3 2 2 p )       A1A1

THEN

attempt to compare coefficients of i, j, and k in their equation or statement to that effect       M1

clear demonstration that the given values satisfy their equation       A1
p = 1, q = 1, r = 4       AG

[5 marks]

b.

attempt at computing AB × AD  (or equivalent)       M1

( 11 10 2 )      A1

area  = | AB × AD | ( = 225 )       (M1)

= 15       A1

[4 marks]

c.

valid attempt to find  OM = ( 1 2 ( a + c ) )       (M1)

( 1 3 2 1 2 )      A1

the equation is

r ( 1 3 2 1 2 ) + t ( 11 10 2 )  or equivalent       M1A1

Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.

[4 marks]

d.

attempt to obtain the equation of the plane in the form ax + by + cz = d       M1

11x + 10y + 2z = 25      A1A1

Note: A1 for right hand side, A1 for left hand side.

[3 marks]

e.

putting two coordinates equal to zero       (M1)

X ( 25 11 , 0 , 0 ) , Y ( 0 , 5 2 , 0 ) , Z ( 0 , 0 , 25 2 )       A1

[2 marks]

f.i.

YZ = ( 5 2 ) 2 + ( 25 2 ) 2      M1

= 325 2 ( = 5 104 4 = 5 26 2 )      A1

[4 marks]

f.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.i.
[N/A]
f.ii.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.12—Vector definitions
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Topic 3— Geometry and trigonometry

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