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Date	November 2017	Marks available	5	Reference code	17N.1.SL.TZ0.S_9
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	S_9	Adapted from	N/A

Question

A line $L$ passes through points ${\text{A}}( - 3,{\text{ }}4,{\text{ }}2)$ and ${\text{B}}( - 1,{\text{ }}3,{\text{ }}3)$ .

The line $L$ also passes through the point ${\text{C}}(3,{\text{ }}1,{\text{ }}p)$ .

Show that $\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ .

[1]

a.i.

Find a vector equation for $L$ .

[2]

a.ii.

Find the value of $p$ .

[5]

The point D has coordinates $({q^2},{\text{ }}0,{\text{ }}q)$ . Given that $\overrightarrow {{\text{DC}}}$ is perpendicular to $L$ , find the possible values of $q$ .

[7]

Markscheme

correct approach A1

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ { - 4} \\ { - 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)$

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ AG N0

[1 mark]

a.i.

any correct equation in the form $r = a + tb$ (any parameter for $t$ )

where $a$ is $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right)$ or $\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)$ and $b$ is a scalar multiple of $\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ A2 N2

eg $\,\,\,\,\,$ $r = \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 3 + 2t} \\ {4 - t} \\ {2 + t} \end{array}} \right)$

Note: Award A1 for the form $a + tb$ , A1 for the form $L = a + tb$ , A0 for the form $r = b + ta$ .

[2 marks]

a.ii.

METHOD 1 – finding value of parameter

valid approach (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)$

one correct equation (not involving $p$ ) (A1)

eg $\,\,\,\,\,$ $- 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1$

correct parameter from their equation (may be seen in substitution) A1

eg $\,\,\,\,\,$ $t = 3,{\text{ }}s = - 2$

correct substitution (A1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}3 - ( - 2)$

$p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)$ A1 N2

METHOD 2 – eliminating parameter

valid approach (M1)

one correct equation (not involving $p$ ) (A1)

eg $\,\,\,\,\,$ $- 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1$

correct equation (with $p$ ) A1

eg $\,\,\,\,\,$ $2 + t = p,{\text{ }}3 - s = p$

correct working to solve for $p$ (A1)

eg $\,\,\,\,\,$ $7 = 2p - 3,{\text{ }}6 = 1 + p$

$p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)$ A1 N2

[5 marks]

valid approach to find $\overrightarrow {{\text{DC}}}$ or $\overrightarrow {{\text{CD}}}$ (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right)$

correct vector for $\overrightarrow {{\text{DC}}}$ or $\overrightarrow {{\text{CD}}}$ (may be seen in scalar product) A1

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2} - 3} \\ { - 1} \\ {q - 5} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right)$

recognizing scalar product of $\overrightarrow {{\text{DC}}}$ or $\overrightarrow {{\text{CD}}}$ with direction vector of $L$ is zero (seen anywhere) (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0,{\text{ }}\overrightarrow {{\text{DC}}} \bullet \overrightarrow {{\text{AC}}} = 0,{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0$