Find $\overrightarrow {AB}$.

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

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 $p = 2$.

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

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

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

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]

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]

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]

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)$

one correct point     A1     N2

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

METHOD 2 (distance between points) 

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

eg$\,\,\,\,\,$${(2s - 8)^2} + {0^2} + {(s - 4)^2} = 5$

solving $5{s^2} - 40s + 75 = 0$ leading to $s = 5$ or $s = 3$     (A1)

one correct point     A1     N2

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

[3 marks]