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Date	November 2018	Marks available	2	Reference code	18N.2.SL.TZ0.S_8
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_8	Adapted from	N/A

Question

Consider the points A(−3, 4, 2) and B(8, −1, 5).

A line L has vector equation $r = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 5} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ 2 \end{array}} \right)$ . The point C (5, $y$ , 1) lies on line L.

Find $\left| {\overrightarrow {{\text{AB}}} } \right|$ .

[2]

a.ii.

Find the value of $y$ .

[3]

b.i.

Show that $\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 8 \\ { - 10} \\ { - 1} \end{array}} \right)$ .

[2]

b.ii.

Find the angle between $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{AC}}}$ .

[5]

Find the area of triangle ABC.

[2]

Markscheme

Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.

correct substitution into formula (A1)

eg $\sqrt {{{11}^2} + {{\left( { - 5} \right)}^2} + {3^2}}$

12.4498

$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {155}$ (exact), 12.4 A1 N2

[2 marks]

a.ii.

valid approach to find $t$ (M1)

eg $\left( {\begin{array}{*{20}{c}} 5 \\ y \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 5} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ 2 \end{array}} \right)$ , $5 = 2 + t$ , $1 = - 5 + 2t$

$t = 3$ (seen anywhere) (A1)

attempt to substitute their parameter into the vector equation (M1)

eg $\left( {\begin{array}{*{20}{c}} 5 \\ y \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 5} \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ 2 \end{array}} \right)$ , $3 \cdot \left( { - 2} \right)$

$y = - 6$ A1 N2

[3 marks]

b.i.

correct approach A1

eg $\left( {\begin{array}{*{20}{c}} 5 \\ { - 6} \\ 1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right)$ , AO + OC, $c - a$

$\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 8 \\ { - 10} \\ { - 1} \end{array}} \right)$ AG N0

Note: Do not award A1 in part (ii) unless answer in part (i) is correct and does not result from working backwards.

[2 marks]

b.ii.

finding scalar product and magnitude (A1)(A1)

scalar product = 11 × 8 + −5 × −10 + 3 × −1 (=135)

$\overrightarrow {\left| {{\text{AC}}} \right|} = \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} \,\,\left( { = \sqrt {165} ,\,\,12.8452} \right)$

evidence of substitution into formula (M1)

eg ${\text{cos}}{\mkern 1mu} \theta = \frac{{11 \times 8 + - 5 \times - 10 + 3 \times - 1}}{{\left| {\overrightarrow {{\text{AB}}} } \right| \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}{\text{,}}\,\,{\text{cos}}{\mkern 1mu} \theta = \frac{{\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} }}{{\sqrt {155} \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}$

correct substitution (A1)

eg ${\text{cos}}\,\theta = \frac{{11 \times 8 + - 5 \times - 10 + 3 \times - 1}}{{\sqrt {155} \times \sqrt {{8^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 1} \right)}^2}} }}$ , ${\text{cos}}\,\theta = \frac{{135}}{{159.921 \ldots }}$ ,

${\text{cos}}\,\theta = 0.844162 \ldots$

0.565795, 32.4177°

${\hat A}$ = 0.566, 32.4° A1 N3

[5 marks]

correct substitution into area formula (A1)

eg $\frac{1}{2} \times \sqrt {155} \times \sqrt {165} \times {\text{sin}}\left( {0.566 \ldots } \right)$ , $\frac{1}{2} \times \sqrt {155 \times 165} \times {\text{sin}}\left( {32.4} \right)$