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.