Find the coordinates of A.

Find $\overrightarrow {{\text{AB}}}$;

Find $\left| {\overrightarrow {{\text{AB}}} } \right|$.

Find $\cos {\rm{B\hat AC}}$.

Hence or otherwise, find the distance between ${P_1}$ and ${P_2}$ two seconds after they leave A.

recognizing $t = 0$ at A     (M1)

A is $(4,{\text{ }} - 1,{\text{ }}3)$     A1     N2

[2 marks]

METHOD 1

valid approach     (M1)

eg$\,\,\,\,\,$$\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} - 1)$

correct approach to find $\overrightarrow {{\text{AB}}}$     (A1)

eg$\,\,\,\,\,$${\text{AO}} + {\text{OB}},{\text{ B}} - {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right)$

$\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)$     A1     N2

METHOD 2

recognizing $\overrightarrow {{\text{AB}}}$ is two times the direction vector     (M1)

correct working     (A1)

eg$\,\,\,\,\,$$\overrightarrow {{\text{AB}}}  = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right)$

$\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)$     A1     N2

[3 marks]

correct substitution     (A1)

eg$\,\,\,\,\,$$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36}$

$\left| {\overrightarrow {{\text{AB}}} } \right| = 6$     A1     N2

[2 marks]

METHOD 1 (vector approach)

valid approach involving $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{AC}}}$     (M1)

eg$\,\,\,\,\,$$\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}}  \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}$

finding scalar product and $\left| {\overrightarrow {{\text{AC}}} } \right|$     (A1)(A1)

scalar product $2(3) + 4(0) - 4(4){\text{ }}( =  - 10)$

$\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)$

substitution of their scalar product and magnitudes into cosine formula     (M1)

eg$\,\,\,\,\,$$\cos {\rm{B\hat AC = }}\frac{{6 + 0 - 16}}{{6\sqrt {{3^2} + {4^2}} }}$

${\text{cos}}\,B\hat AC =  - \frac{{10}}{{30}}\left( { =  - \frac{1}{3}} \right)$     A1     N2

METHOD 2 (triangle approach)

valid approach involving cosine rule     (M1)

eg$\,\,\,\,\,$$\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} - {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}$

finding lengths AC and BC     (A1)(A1)

${\text{AC}} = 5,{\text{ BC}} = 9$

substitution of their lengths into cosine formula     (M1)

eg$\,\,\,\,\,$$\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} - {9^2}}}{{2 \times 5 \times 6}}$

$\cos {\rm{B\hat AC}} =  - \frac{{20}}{{60}}{\text{ }}\left( { =  - \frac{1}{3}} \right)$     A1     N2

[5 marks]

Note:     Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

METHOD 1 (using cosine rule)

recognizing need to find BC     (M1)

choosing cosine rule     (M1)

eg$\,\,\,\,\,$${c^2} = {a^2} + {b^2} - 2ab\cos {\text{C}}$

correct substitution into RHS     A1

eg$\,\,\,\,\,$${\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} - 2(6)(5)\left( { - \frac{1}{3}} \right),{\text{ }}36 + 25 + 20$

distance is 9     A1     N2

METHOD 2 (finding magnitude of $\overrightarrow {BC}$) 

recognizing need to find BC     (M1)

valid approach     (M1)

eg$\,\,\,\,\,$attempt to find $\overrightarrow {{\rm{OB}}}$ or $\overrightarrow {{\rm{OC}}}$, $\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right)$ or $\overrightarrow {{\rm{OC}}}  = \left( {\begin{array}{*{20}{c}} 7 \\ { - 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}}  + \overrightarrow {{\rm{AC}}}$

correct working     A1

eg$\,\,\,\,\,$$\overrightarrow {{\rm{BC}}}  = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}}  = \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \\ { - 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81}$

distance is 9     A1     N2

METHOD 3 (finding coordinates and using distance formula)

recognizing need to find BC     (M1)

valid approach     (M1)

eg$\,\,\,\,\,$attempt to find coordinates of B or C, ${\text{B}}(6,{\text{ }}3,{\text{ }} - 1)$ or ${\text{C}}(7,{\text{ }} - 1,{\text{ }}7)$

correct substitution into distance formula     A1

eg$\,\,\,\,\,$${\text{BC}} = \sqrt {{{(6 - 7)}^2} + {{\left( {3 - ( - 1)} \right)}^2} + {{( - 1 - 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81}$

distance is 9     A1     N2

[4 marks]