Find $\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}$.

Hence find the area of the triangle ABC.

$\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}   1 \\   0 \\   5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   0 \\   { - 2} \\   2 \end{array}} \right)$, $\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}}   2 \\   { - 1} \\   4 \end{array}} \right) - \left( {\begin{array}{*{20}{c}}   1 \\   2 \\   3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}   1 \\   { - 3} \\   1 \end{array}} \right)$     A1A1

Note: Award the above marks if the components are seen in the line below.

$\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}   i&j&k \\   0&{ - 2}&2 \\   1&{ - 3}&1 \end{array}} \right| = \left( {\begin{array}{*{20}{c}}   4 \\   2 \\   2 \end{array}} \right)$     (M1)A1

[4 marks]

area $= \frac{1}{2}\left| {\left( {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right)} \right|$     (M1)

$= \frac{1}{2}\sqrt {{4^2} + {2^2} + {2^2}}  = \frac{1}{2}\sqrt {24} {\text{ }}\left( { = \sqrt 6 } \right)$     A1

Note: Award M0A0 for attempts that do not involve the answer to (a).

[2 marks]

Candidates showed a good understanding of the vector techniques required in this question.

Candidates showed a good understanding of the vector techniques required in this question.