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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	Show that	Question number	S_8	Adapted from	N/A

Question

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

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

Find $\overrightarrow {AB}$ .

[2]

a.i.

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

[2]

a.ii.

A second line ${L_2}$ , has equation r = $\left( {\begin{array}{*{20}{c}} 1 \\ {13} \\ { - 14} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right)$ .

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

[3]

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

[5]

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

[2]

d.i.

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

[3]

d.ii.

Markscheme

valid approach (M1)

eg $A - B,\,\, - \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 8 \hfill \\ \end{gathered} \right) + \left( \begin{gathered} 3 \hfill \\ 5 \hfill \\ 2 \hfill \\ \end{gathered} \right)$

$\overrightarrow {AB} = \left( \begin{gathered} 3 \hfill \\ 4 \hfill \\ - 6 \hfill \\ \end{gathered} \right)$ A1 N2

[2 marks]

a.i.

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

where a is $\left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 8 \hfill \\ \end{gathered} \right)$ or $\left( \begin{gathered} 3 \hfill \\ 5 \hfill \\ 2 \hfill \\ \end{gathered} \right)$ , and b is a scalar multiple of $\left( \begin{gathered} 3 \hfill \\ 4 \hfill \\ - 6 \hfill \\ \end{gathered} \right)$

eg $\,\,\,\,\,$ r = $\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)$ , r = $\left( {\begin{array}{*{20}{c}} {3 + 3t} \\ {5 + 4t} \\ {2 - 6t} \end{array}} \right)$ , 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 $\,\,\,\,\,$ $a \bullet b = 0$

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

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right) = 0$

correct working/equation A1

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

$p = 2$ AG N0

[3 marks]

valid approach (M1)

eg $\,\,\,\,\,$ ${L_1} = \left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ z \end{array}} \right),{\text{ }}{L_1} = {L_2}$

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

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

one correct value A1

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

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

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

$z = - 10$ A1 N3

[5 marks]

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

$\frac{1}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)\,\,\,\,\,\left( {{\text{accept}}\left( {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}} \\ {\frac{0}{{\sqrt 5 }}} \\ {\frac{1}{{\sqrt 5 }}} \end{array}} \right)} \right)$ A1 N2

[2 marks]

d.i.

METHOD 1 (using unit vector)

valid approach (M1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) \pm \sqrt 5 \hat d$

correct working (A1)

eg $\,\,\,\,\,$ $\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)$