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Date May 2018 Marks available 2 Reference code 18M.2.SL.TZ2.S_8
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number S_8 Adapted from N/A

Question

Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.

Let  PR = 6i − j + 3k.

Find  PQ .

[2]
a.i.

Find | PQ | .

[2]
a.ii.

Find the angle between PQ and PR.

[4]
b.

Find the area of triangle PQR.

[2]
c.

Hence or otherwise find the shortest distance from R to the line through P and Q.

[3]
d.

Markscheme

valid approach      (M1)

eg   (7, 4, 9) − (3, 2, 5)  A − B

PQ = 4i + 2j + 4k  ( = ( 4 2 4 ) )      A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula      (A1)
eg   4 2 + 2 2 + 4 2

| PQ | = 6      A1 N2

[2 marks]

a.ii.

finding scalar product and magnitudes      (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR =  36 + 1 + 9 = ( 6.782 )

correct substitution of their values to find cos  Q P R      M1

eg  cos  Q P R  =  24 2 + 12 ( 6 ) × ( 46 ) , 0.8355

0.581746

Q P R = 0.582 radians  or  Q P R = 33.3°     A1 N3

[4 marks]

b.

correct substitution (A1)
eg     1 2 × | PQ | × | PR | × sin P , 1 2 × 6 × 46 × sin 0.582

area is 11.2 (sq. units)      A1 N2

[2 marks]

c.

recognizing shortest distance is perpendicular distance from R to line through P and Q      (M1)

eg  sketch, height of triangle with base  [ PQ ] , 1 2 × 6 × h , sin 33.3 = h 46

correct working      (A1)

eg   1 2 × 6 × d = 11.2 , | PR | × sin P , 46 × sin 33.3

3.72677

distance = 3.73  (units)    A1 N2

[3 marks]

d.

Examiners report

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

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
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Topic 3— Geometry and trigonometry

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