(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]