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Date May 2018 Marks available 2 Reference code 18M.2.sl.TZ2.8
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 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 (=(424))     A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula      (A1)
eg  42+22+42

|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 QPR     M1

eg  cos QPR = 242+12(6)×(46),0.8355

0.581746

QPR = 0.582 radians  or QPR = 33.3°     A1 N3

[4 marks]

b.

correct substitution (A1)
eg    12×|PQ|×|PR|×sinP,12×6×46×sin0.582

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],12×6×h,sin33.3=h46

correct working      (A1)

eg  12×6×d=11.2,|PR|×sinP,46×sin33.3

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 4 - Vectors » 4.1 » Vectors as displacements in the plane and in three dimensions.

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