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]