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]