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Date	November 2017	Marks available	1	Reference code	17N.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

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

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

Show that $- - \to AB = ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ .

[1]

a.i.

Find a vector equation for $L$ $L$ .

[2]

a.ii.

Find the value of $p$ $p$ .

[5]

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

[7]

Markscheme

correct approach A1

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} - 1 33 \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} - 3 42 \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} 3 - 4 - 2 \end{matrix} ⎞ ⎟ ⎠ + ⎛ ⎜ ⎝ \begin{matrix} - 1 33 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ { - 4} \\ { - 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)$

$- - \to AB = ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ AG N0

[1 mark]

a.i.

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

where $a$ $a$ is $⎛ ⎜ ⎝ \begin{matrix} - 3 42 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right)$ or $⎛ ⎜ ⎝ \begin{matrix} - 1 33 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)$ and $b$ $b$ is a scalar multiple of $⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)$ A2 N2

eg $\,\,\,\,\,$ $r = ⎛ ⎜ ⎝ \begin{matrix} - 3 42 \end{matrix} ⎞ ⎟ ⎠ + t ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠, (x, y, z) = (- 1, 3, 3) + s (- 2, 1, - 1), r = ⎛ ⎜ ⎝ \begin{matrix} - 3 + 2 t 4 - t 2 + t \end{matrix} ⎞ ⎟ ⎠$ $r = \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 3 + 2t} \\ {4 - t} \\ {2 + t} \end{array}} \right)$

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

[2 marks]

a.ii.

METHOD 1 – finding value of parameter

valid approach (M1)

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} - 3 42 \end{matrix} ⎞ ⎟ ⎠ + t ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} 31 p \end{matrix} ⎞ ⎟ ⎠, (- 1, 3, 3) + s (- 2, 1, - 1) = (3, 1, p)$ $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)$

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

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

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

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

correct substitution (A1)

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} - 3 42 \end{matrix} ⎞ ⎟ ⎠ + 3 ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠ = ⎛ ⎜ ⎝ \begin{matrix} 31 p \end{matrix} ⎞ ⎟ ⎠, 3 - (- 2)$ $\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}3 - ( - 2)$

$p = 5 ⎛ ⎜ ⎝ accept ⎛ ⎜ ⎝ \begin{matrix} 315 \end{matrix} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠$ $p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)$ A1 N2

METHOD 2 – eliminating parameter

valid approach (M1)

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

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

correct equation (with $p$ $p$ ) A1

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

correct working to solve for $p$ $p$ (A1)

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

[5 marks]

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

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 315 \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} q^{2} 0 q \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} q^{2} 0 q \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} 315 \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} q^{2} 0 q \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} 31 p \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right)$

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

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 3 - q^{2} 1 5 - q \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} q^{2} - 3 - 1 q - 5 \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} 3 - q^{2} 1 p - q \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2} - 3} \\ { - 1} \\ {q - 5} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right)$

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

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 3 - q^{2} 1 p - q \end{matrix} ⎞ ⎟ ⎠ ∙ ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠ = 0, - - \to DC ∙ - - \to AC = 0, ⎛ ⎜ ⎝ \begin{matrix} 3 - q^{2} 1 5 - q \end{matrix} ⎞ ⎟ ⎠ ∙ ⎛ ⎜ ⎝ \begin{matrix} 2 - 1 1 \end{matrix} ⎞ ⎟ ⎠ = 0$ $\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0,{\text{ }}\overrightarrow {{\text{DC}}} \bullet \overrightarrow {{\text{AC}}} = 0,{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0$