Show that $\overrightarrow {{\text{AB}}}  =$  $\left( \begin{array}{c} - 1\\0\\1\end{array} \right)$

Hence, write down a direction vector for ${L_1}$;

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

Find the coordinates of P.

Write down a direction vector for ${L_2}$.

Hence, find the angle between ${L_1}$ and ${L_2}$.

correct approach     A1

eg   $\left( \begin{array}{c}1\\1\\5\end{array} \right) - \left( \begin{array}{l}2\\1\\4\end{array} \right)$, ${\text{AO}} + {\text{OB}}$, $b - a$

$\overrightarrow {{\text{AB}}}  =$  $\left( \begin{array}{c} - 1\\0\\1\end{array} \right)$    AG     N0

[1 mark]

correct vector (or any multiple)     A1     N1

eg     d =  $\left( \begin{array}{c} - 1\\0\\1\end{array} \right)$

[1 mark]

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

where a is $\left( \begin{array}{c}2\\1\\4\end{array} \right)$ or $\left( \begin{array}{c}1\\1\\5\end{array} \right)$ , and b is a scalar multiple of  $\left( \begin{array}{c} - 1\\0\\1\end{array} \right)$     A2     N2

eg   r = $\left( \begin{array}{c}1\\1\\5\end{array} \right) + t\left( \begin{array}{c} - 1\\0\\1\end{array} \right),\left( \begin{array}{c}x\\y\\z\end{array} \right) = \left( \begin{array}{c}2 - s\\1\\4 + s\end{array} \right)$

Note:     Award A1 for a + tb, A1 for ${L_1}$ = a + tb, A0 for r = b + ta.

[2 marks]

valid approach     (M1)

eg     ${r_1} = {r_2}$, $\left( \begin{array}{c}2\\1\\4\end{array} \right) + t\left( \begin{array}{c} - 1\\0\\1\end{array} \right) = \left( \begin{array}{c}4\\7\\ - 4\end{array} \right) + s\left( \begin{array}{c}0\\ - 1\\1\end{array} \right)$

one correct equation in one parameter     A1

eg     $2 - t = 4, 1 = 7 - s, 1 - t = 4$

attempt to solve     (M1)

eg     $2 - 4 = t, s = 7 - 1, t = 1 - 4$

one correct parameter     A1

eg     $t = -2, s = 6, t = -3$,

attempt to substitute their parameter into vector equation     (M1)

eg     $\left( \begin{array}{c}4\\7\\ - 4\end{array} \right) + 6\left( \begin{array}{c}0\\ - 1\\1\end{array} \right)$

P(4, 1, 2)   (accept position vector)     A1     N2

[6 marks]

correct direction vector for ${L_2}$     A1     N1

eg     $\left( \begin{array}{c}0\\-1\\ 1\end{array} \right)$, $\left( \begin{array}{c}0\\2\\ - 2\end{array} \right)$

[1 mark]

correct scalar product and magnitudes for their direction vectors     (A1)(A1)(A1)

scalar product $= 0 \times  - 1 +  - 1 \times 0 + 1 \times 1{\text{ }}( = 1)$

magnitudes $= \sqrt {{0^2} + {{( - 1)}^2} + {1^2}} ,{\text{ }}\sqrt { - {1^2} + {0^2} + {1^2}} \left( {\sqrt 2 ,{\text{ }}\sqrt 2 } \right)$

attempt to substitute their values into formula     M1

eg   $\frac{{0 + 0 + 1}}{{\left( {\sqrt {{0^2} + {{( - 1)}^2} + {1^2}} } \right) \times \left( {\sqrt { - {1^2} + {0^2} + {1^2}} } \right)}},{\text{ }}\frac{1}{{\sqrt 2  \times \sqrt 2 }}$

correct value for cosine, $\frac{1}{2}$     A1

angle is $\frac{\pi }{3}{\text{ }}( = {60^ \circ })$     A1     N1

[6 marks]