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]