Find $\overrightarrow {{\text{BR}}}$ in terms of ${{a}}$, ${{b}}$ and ${{c}}$.

(i)     Find a vector equation of the line that passes through $B$ and $R$ in terms of ${{a}}$, ${{b}}$ and ${{c}}$ and a parameter $\lambda$.

(ii)     Find a vector equation of the line that passes through $A$ and $Q$ in terms of ${{a}}$, ${{b}}$ and ${{c}}$ and a parameter $\mu$.

(iii)     Hence show that $\overrightarrow {{\text{OG}}}  = \frac{1}{3}({{a}} + {{b}} + {{c}})$ given that $G$ is the point where [$BR$] and [$AQ$] intersect.

Show that the line segment [$CP$] also includes the point $G$.

The coordinates of the points $A$, $B$ and $C$ are $(1,{\text{ }}3,{\text{ }}1)$, $(3,{\text{ }}7,{\text{ }} - 5)$ and $(2,{\text{ }}2,{\text{ }}1)$ respectively.

A point $X$ is such that [$GX$] is perpendicular to the plane $ABC$.

Given that the tetrahedron $ABCX$ has volume ${\text{12 unit}}{{\text{s}}^{\text{3}}}$, find possible coordinates

of $X$.

$\overrightarrow {{\text{BR}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}}  + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)$     (M1)

$= ({{a}} - {{b}}) + \frac{1}{2}({{c}} - {{a}})$

$= \frac{1}{2}{{a}} - {{b}} + \frac{1}{2}{{c}}$     A1

[2 marks]

(i)     ${{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} - {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 - \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)$     A1A1

Note:     Award A1A0 if the ${\text{r}} =$ is omitted in an otherwise correct expression/equation.

Do not penalise such an omission more than once.

(ii)     $\overrightarrow {{\text{AQ}}}  =  - {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}$     (A1)

${{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { - {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 - \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)$     A1

Note:     Accept the use of the same parameter in (i) and (ii).

(iii)     when $\overrightarrow {{\text{AQ}}}$ and $\overrightarrow {{\text{BP}}}$ intersect we will have ${{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}$     (M1)

Note:     If the same parameters are used for both equations, award at most M1M1A0A0M1.

$\frac{\lambda }{2}{{a}} + (1 - \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 - \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}$

attempt to equate the coefficients of the vectors ${{a}}$, ${{b}}$ and ${{c}}$     M1

$\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 - \mu } \\ {1 - \lambda  = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}$     (A1)

$\lambda  = \frac{2}{3}$ or $\mu  = \frac{2}{3}$     A1

substituting parameters back into one of the equations     M1

$\overrightarrow {{\text{OG}}}  = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 - \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})$     AG

Note:     Accept solution by verification.

[9 marks]

$\overrightarrow {{\text{CP}}}  = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} - {{c}}$     (M1)A1

so we have that ${{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} - {{c}}} \right)$ and when $\beta  = \frac{2}{3}$ the line passes through

the point $G$ (ie, with position vector $\frac{1}{3}({{a}} + {{b}} + {{c}})$)     R1

hence [$AQ$], [$BR$] and [$CP$] all intersect in $G$     AG

[3 marks]

$\overrightarrow {{\text{OG}}}  = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { - 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 1} \end{array}} \right)$     A1

Note:     This independent mark for the vector may be awarded wherever the vector is calculated.

$\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 6} \\ { - 6} \\ { - 6} \end{array}} \right)$     M1A1

$\overrightarrow {{\text{GX}}}  = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)$     (M1)

volume of Tetrahedron given by $\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}$

$= \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12$     (M1)(A1)

Note:     Accept alternative methods, for example the use of a scalar triple product.

$= \frac{1}{6}\sqrt {{{( - 6)}^2} + {{( - 6)}^2} + {{( - 6)}^2}}  \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}}  = 12$     (A1)

$= \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3  = 12$

$\Rightarrow |\alpha | = 4$     A1

Note:     Condone absence of absolute value.

this gives us the position of $X$ as $\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)$

${\text{X}}(6,{\text{ }}8,{\text{ }}3)$ or $( - 2,{\text{ }}0,{\text{ }} - 5)$     A1

Note:     Award A1 for either result.

[9 marks]

Total [23 marks]