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Date	November 2019	Marks available	8	Reference code	19N.1.SL.TZ0.S_9
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Show that	Question number	S_9	Adapted from	N/A

Question

The points ${\text{A}}$ and ${\text{B}}$ have position vectors $\left( {\begin{array}{*{20}{c}} { - 2} \\ 4 \\ { - 4} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} 6 \\ 8 \\ 0 \end{array}} \right)$ respectively.

Point ${\text{C}}$ has position vector $\left( {\begin{array}{*{20}{c}} { - 1} \\ k \\ 0 \end{array}} \right)$ . Let ${\text{O}}$ be the origin.

Find, in terms of $k$ ,

$\overrightarrow {{\text{OA}}} \bullet \overrightarrow {{\text{OC}}}$ .

[2]

a.i.

$\overrightarrow {{\text{OB}}} \bullet \overrightarrow {{\text{OC}}}$ .

[1]

a.ii.

Given that ${\text{A}}\widehat {\text{O}}{\text{C}} = {\text{B}}\widehat {\text{O}}{\text{C}}$ , show that $k = 7$ .

[8]

Calculate the area of triangle ${\text{AOC}}$ .

[6]

Markscheme

correct substitution into either $\overrightarrow {{\text{OA}}} \bullet \overrightarrow {{\text{OC}}}$ or into $\overrightarrow {{\text{OB}}} \bullet \overrightarrow {{\text{OC}}}$ (in (ii)) (A1)

eg $- 2 \times \left( { - 1} \right) + 4 \times k$ , $6 \times \left( { - 1} \right) + 8 \times k$

correct expression A1 N1

eg $2 + 4k$ , $4k + 2$

[2 marks]

a.i.

correct expression A1 N1

eg $8k - 6$ , $- 6 + 8k$

[1 mark]

a.ii.

finding magnitudes (seen anywhere) A1A1

eg $\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} \,\,\left( { = 6} \right)$ , $\sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {0^2}} \,\,\left( { = 10} \right)$

correct substitution of their values into formula for angle ${\text{AOC}}$ (A1)

eg ${\text{cos}}\,\theta = \frac{{2 + 4k}}{{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} \left| {\overrightarrow {{\text{OC}}} } \right|}}$

correct substitution of their values into formula for angle ${\text{BOC}}$ (A1)

eg ${\text{cos}}\,\theta = \frac{{8k - 6}}{{\sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {0^2}} \left| {\overrightarrow {{\text{OC}}} } \right|}}$

recognizing that ${\text{cos}}\,{\text{A}}\widehat {\text{O}}{\text{C}} = {\text{cos}}\,{\text{B}}\widehat {\text{O}}{\text{C}}$ (seen anywhere) (M1)

eg $\frac{{2 + 4k}}{{\left| {\overrightarrow {{\text{OC}}} } \right|\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{8k - 6}}{{\left| {\overrightarrow {{\text{OC}}} } \right|\sqrt {{6^2} + {{\left( 8 \right)}^2} + {0^2}} }}$ , $\frac{{2 + 4k}}{{6\sqrt {1 + {k^2}} }} = \frac{{8k - 6}}{{10\sqrt {1 + {k^2}} }}$

correct working (without radicals) (A2)

eg $10\left( {2 + 4k} \right) = 6\left( {8k - 6} \right)$ , $11{k^2} - 79k + 14 = 0$

correct working clearly leading to the required answer A1

eg $20 + 36 = 48k$ , $56 = 8k$ , $k = 7$ and $k = \frac{2}{{11}}$ , $\left( {k - 7} \right)\left( {11k - 2} \right) = 0$

$k = 7$ AG N0

[8 marks]

finding magnitude of $\overrightarrow {{\text{OC}}}$ (seen anywhere) A1

eg $\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}}$ , $\sqrt {50}$

valid attempt to find ${\text{cos}}\,\theta$ (M1)

eg ${\text{cos}}\,\theta = \frac{{2 + 28}}{{6\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}} }}$ , ${\text{cos}}\,\theta = \frac{{56 - 6}}{{10\sqrt {{{\left( { - 1} \right)}^2} + {7^2} + {0^2}} }}$ , ${\left( {\sqrt {26} } \right)^2} = {6^2} + {\left( {\sqrt {50} } \right)^2} - 2\left( 6 \right)\sqrt {50} \,{\text{cos}}\,\theta$

finding ${\text{cos}}\,\theta$ A1

eg ${\text{cos}}\,\theta = \frac{5}{{\sqrt {50} }}\,\,\,\left( { = \frac{1}{{\sqrt 2 }}} \right)$

valid approach to find ${\text{sin}}\,\theta$ (seen anywhere) (M1)

eg $\theta = \frac{\pi }{4}$ , ${\text{sin}}\,\theta = {\text{cos}}\,\theta$ , ${\text{sin}}\,\theta = \sqrt {1 - \frac{{25}}{{50}}}$ , ${\text{sin}}\,\theta = \sqrt {1 - {\text{co}}{{\text{s}}^2}\,\theta }$ , ${\text{sin}}\,\theta = \frac{{\sqrt 2 }}{2}$