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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 ${\mathop {{\text{PR}}}\limits^ \to }$ = 6i − j + 3k.

Find $\mathop {{\text{PQ}}}\limits^ \to$ .

[2]

a.i.

Find $\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|$ .

[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

$\mathop {{\text{PQ}}}\limits^ \to =$ 4i + 2j + 4k $\left( { = \left( \begin{gathered} 4 \hfill \\ 2 \hfill \\ 4 \hfill \\ \end{gathered} \right)} \right)$ A1 N2

[2 marks]

a.i.

correct substitution into magnitude formula (A1)
eg $\sqrt {{4^2} + {2^2} + {4^2}}$

$\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6$ 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 = $\sqrt {36 + 1 + 9} = \left( {6.782} \right)$

correct substitution of their values to find cos ${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$ M1

eg cos ${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 - 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355$

0.581746

${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$ = 0.582 radians or ${\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}$ = 33.3° A1 N3

[4 marks]

b.

correct substitution (A1)
eg $\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582$

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 $\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}$

correct working (A1)

eg $\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ$

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 and vector products

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