(i)     Find $\overrightarrow {{\text{AB}}}$.

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

Show that the coordinates of C are $( - 2,{\text{ }}1,{\text{ }}3)$.

Write down an expression in terms of $\theta$ for

(i)     angle ADB;

(ii)     area of triangle ABD.

Given that $\frac{{{\text{area }}\Delta {\text{ABD}}}}{{{\text{area }}\Delta {\text{ACD}}}} = 3$, show that $\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}$.

Hence or otherwise, find the coordinates of point D.

(i)     valid approach to find $\overrightarrow {{\text{AB}}}$

eg$\,\,\,\,\,$$\overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OA}}} ,{\text{ }}\left( {\begin{array}{*{20}{c}} {4 - ( - 1)} \\ {1 - 0} \\ {3 - 4} \end{array}} \right)$

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 5 \\ 1 \\ { - 1} \end{array}} \right)$    A1     N2

(ii)     valid approach to find $\left| {\overrightarrow {{\text{AB}}} } \right|$     (M1)

eg$\,\,\,\,\,$$\sqrt {{{(5)}^2} + {{(1)}^2} + {{( - 1)}^2}}$

$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {27}$     A1     N2

[4 marks]

correct approach     A1

eg$\,\,\,\,\,$$\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)$

$C$ has coordinates $( - 2,{\text{ }}1,{\text{ }}3)$     AG     N0

[1 mark]

(i)     ${\rm{A\hat DB}} = \pi  - \theta ,{\rm{ \hat D}} = 180 - \theta$     A1     N1

(ii)     any correct expression for the area involving $\theta$     A1     N1

eg$\,\,\,\,\,$${\text{area}} = \frac{1}{2} \times {\text{AD}} \times {\text{BD}} \times \sin (180 - \theta ),{\text{ }}\frac{1}{2}ab\sin \theta ,{\text{ }}\frac{1}{2}\left| {\overrightarrow {{\text{DA}}} } \right|\left| {\overrightarrow {{\text{DB}}} } \right|\sin (\pi  - \theta )$

[2 marks]

METHOD 1 (using sine formula for area)

correct expression for the area of triangle ACD (seen anywhere)     (A1)

eg$\,\,\,\,\,$$\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta$

correct equation involving areas     A1

eg$\,\,\,\,\,$$\frac{{\frac{1}{2}{\text{AD}} \times {\text{BD}} \times \sin (\pi  - \theta )}}{{\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta }} = 3$

recognizing that $\sin (\pi  - \theta ) = \sin \theta$ (seen anywhere)     (A1)

$\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3$ (seen anywhere)     (A1)

correct approach using ratio     A1

eg$\,\,\,\,\,$$3\overrightarrow {{\text{DC}}}  + \overrightarrow {{\text{DC}}}  = \overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{BC}}}  = 4\overrightarrow {{\text{DC}}}$

correct ratio $\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}$     AG     N0

METHOD 2 (Geometric approach)

recognising $\Delta {\text{ABD}}$ and $\Delta {\text{ACD}}$ have same height     (A1)

eg$\,\,\,\,\,$use of $h$ for both triangles, $\frac{{\frac{1}{2}{\text{BD}} \times h}}{{\frac{1}{2}{\text{CD}} \times h}} = 3$

correct approach     A2

eg$\,\,\,\,\,$${\text{BD}} = 3x$ and ${\text{DC}} = x,{\text{ }}\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3$

correct working     A2

eg$\,\,\,\,\,$${\text{BC}} = 4x,{\text{ BD}} + {\text{DC}} = 4{\text{DC, }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3x}}{{4x}},{\text{ }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3{\text{DC}}}}{{4{\text{DC}}}}$

$\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}$    AG     N0

[5 marks]

correct working (seen anywhere)     (A1)

eg$\,\,\,\,\,$$\overrightarrow {{\text{BD}}} = \frac{3}{4}\overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{CD}}} = \frac{1}{4}\overrightarrow {{\text{CB}}}$

valid approach (seen anywhere)     (M1)

eg$\,\,\,\,\,$$\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right)$

correct working to find $x$-coordinate     (A1)

eg$\,\,\,\,\,$$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right) + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}x = 4 + \frac{3}{4}( - 6),{\text{ }} - 2 + \frac{1}{4}(6)$

D is $\left( { - \frac{1}{2},{\text{ }}1,{\text{ }}3} \right)$     A1     N3

[4 marks]