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Date	May 2017	Marks available	3	Reference code	17M.1.SL.TZ1.S_8
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find and Hence or otherwise	Question number	S_8	Adapted from	N/A

Question

A line $L_{1}$ ${L_1}$ passes through the points $A (0, 1, 8)$ ${\text{A}}(0,{\text{ }}1,{\text{ }}8)$ and $B (3, 5, 2)$ ${\text{B}}(3,{\text{ }}5,{\text{ }}2)$ .

Given that $L_{1}$ ${L_1}$ and $L_{2}$ ${L_2}$ are perpendicular, show that $p = 2$ $p = 2$ .

Find $- - \to A B$ $\overrightarrow {AB}$ .

[2]

a.i.

Hence, write down a vector equation for $L_{1}$ ${L_1}$ .

[2]

a.ii.

A second line $L_{2}$ ${L_2}$ , has equation r = $⎛ ⎜ ⎝ \begin{matrix} 113 - 14 \end{matrix} ⎞ ⎟ ⎠ + s ⎛ ⎜ ⎝ \begin{matrix} p 01 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 1 \\ {13} \\ { - 14} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right)$ .

Given that $L_{1}$ ${L_1}$ and $L_{2}$ ${L_2}$ are perpendicular, show that $p = 2$ $p = 2$ .

[3]

The lines $L_{1}$ ${L_1}$ and $L_{1}$ ${L_1}$ intersect at $C (9, 13, z)$ $C(9,{\text{ }}13,{\text{ }}z)$ . Find $z$ $z$ .

[5]

Find a unit vector in the direction of $L_{2}$ ${L_2}$ .

[2]

d.i.

Hence or otherwise, find one point on $L_{2}$ ${L_2}$ which is $\sqrt{5}$ $\sqrt 5$ units from C.

[3]

d.ii.

Markscheme

valid approach (M1)

eg $A - B, - ⎛ ⎝ \begin{matrix} 018 \end{matrix} ⎞ ⎠ + ⎛ ⎝ \begin{matrix} 352 \end{matrix} ⎞ ⎠$ $A - B,\,\, - \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 8 \hfill \\ \end{gathered} \right) + \left( \begin{gathered} 3 \hfill \\ 5 \hfill \\ 2 \hfill \\ \end{gathered} \right)$

$- - \to A B = ⎛ ⎝ \begin{matrix} 34 - 6 \end{matrix} ⎞ ⎠$ $\overrightarrow {AB} = \left( \begin{gathered} 3 \hfill \\ 4 \hfill \\ - 6 \hfill \\ \end{gathered} \right)$ A1 N2

[2 marks]

a.i.

any correct equation in the form r = a + tb (any parameter for $t$ $t$ ) A2 N2

where a is $⎛ ⎝ \begin{matrix} 018 \end{matrix} ⎞ ⎠$ $\left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 8 \hfill \\ \end{gathered} \right)$ or $⎛ ⎝ \begin{matrix} 352 \end{matrix} ⎞ ⎠$ $\left( \begin{gathered} 3 \hfill \\ 5 \hfill \\ 2 \hfill \\ \end{gathered} \right)$ , and b is a scalar multiple of $⎛ ⎝ \begin{matrix} 34 - 6 \end{matrix} ⎞ ⎠$ $\left( \begin{gathered} 3 \hfill \\ 4 \hfill \\ - 6 \hfill \\ \end{gathered} \right)$

eg $\,\,\,\,\,$ r = $⎛ ⎜ ⎝ \begin{matrix} 018 \end{matrix} ⎞ ⎟ ⎠ + t ⎛ ⎜ ⎝ \begin{matrix} 34 - 6 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)$ , r = $⎛ ⎜ ⎝ \begin{matrix} 3 + 3 t 5 + 4 t 2 - 6 t \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} {3 + 3t} \\ {5 + 4t} \\ {2 - 6t} \end{array}} \right)$ , r = j + 8k + t(3i + 4j – 6k)

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.

valid approach (M1)

eg $\,\,\,\,\,$ $a ∙ b = 0$ $a \bullet b = 0$

choosing correct direction vectors (may be seen in scalar product) A1

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 34 - 6 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)$ and $⎛ ⎜ ⎝ \begin{matrix} p 01 \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} 34 - 6 \end{matrix} ⎞ ⎟ ⎠ ∙ ⎛ ⎜ ⎝ \begin{matrix} p 01 \end{matrix} ⎞ ⎟ ⎠ = 0$ $\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right) = 0$

correct working/equation A1

eg $\,\,\,\,\,$ $3 p - 6 = 0$ $3p - 6 = 0$

$p = 2$ $p = 2$ AG N0

[3 marks]

valid approach (M1)

eg $\,\,\,\,\,$ $L_{1} = ⎛ ⎜ ⎝ \begin{matrix} 913 z \end{matrix} ⎞ ⎟ ⎠, L_{1} = L_{2}$ ${L_1} = \left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ z \end{array}} \right),{\text{ }}{L_1} = {L_2}$

one correct equation (must be different parameters if both lines used) (A1)

eg $\,\,\,\,\,$ $3 t = 9, 1 + 2 s = 9, 5 + 4 t = 13, 3 t = 1 + 2 s$ $3t = 9,{\text{ }}1 + 2s = 9,{\text{ }}5 + 4t = 13,{\text{ }}3t = 1 + 2s$

one correct value A1

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

valid approach to substitute their $t$ $t$ or $s$ $s$ value (M1)

eg $\,\,\,\,\,$ $8 + 3 (- 6), - 14 + 4 (1)$ $8 + 3( - 6),{\text{ }} - 14 + 4(1)$

$z = - 10$ $z = - 10$ A1 N3

[5 marks]

$∣ ∣ \to d ∣ ∣ = \sqrt{2^{2} + 1} (= \sqrt{5})$ $\left| {\vec d} \right| = \sqrt {{2^2} + 1} \,\,\left( { = \sqrt 5 } \right)$ (A1)

$\frac{1}{\sqrt{5}} ⎛ ⎜ ⎝ \begin{matrix} 201 \end{matrix} ⎞ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ accept ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} \frac{2}{\sqrt{5}} \frac{0}{\sqrt{5}} \frac{1}{\sqrt{5}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ $\frac{1}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)\,\,\,\,\,\left( {{\text{accept}}\left( {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}} \\ {\frac{0}{{\sqrt 5 }}} \\ {\frac{1}{{\sqrt 5 }}} \end{array}} \right)} \right)$ A1 N2

[2 marks]

d.i.

METHOD 1 (using unit vector)

valid approach (M1)

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 913 - 10 \end{matrix} ⎞ ⎟ ⎠ \pm \sqrt{5}^d$ $\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) \pm \sqrt 5 \hat d$

correct working (A1)

eg $\,\,\,\,\,$ $⎛ ⎜ ⎝ \begin{matrix} 913 - 10 \end{matrix} ⎞ ⎟ ⎠ + ⎛ ⎜ ⎝ \begin{matrix} 201 \end{matrix} ⎞ ⎟ ⎠, ⎛ ⎜ ⎝ \begin{matrix} 913 - 10 \end{matrix} ⎞ ⎟ ⎠ - ⎛ ⎜ ⎝ \begin{matrix} 201 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)$

one correct point A1 N2

eg $\,\,\,\,\,$ $(11, 13, - 9), (7, 13, - 11)$ $(11,{\text{ }}13,{\text{ }} - 9),{\text{ }}(7,{\text{ }}13,{\text{ }} - 11)$

METHOD 2 (distance between points)

attempt to use distance between $(1 + 2 s, 13, - 14 + s)$ $(1 + 2s,{\text{ }}13,{\text{ }} - 14 + s)$ and $(9, 13, - 10)$ $(9,{\text{ }}13,{\text{ }} - 10)$ (M1)