The points $A$, $B$ and $C$ have coordinates $A(1,0,2)$, $B(3,1,1)$ and $C(2,4,0)$.
(a) Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$. [2]
(b) Find $\overrightarrow{AB}\times\overrightarrow{AC}$. [2]
(c) Hence find the area of triangle $ABC$. [2]
- $\overrightarrow{AB}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$A1
- $\overrightarrow{AC}=\begin{pmatrix}1\\4\\-2\end{pmatrix}$A1
- Cross product with at least two components correctM1
- $\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}2\\3\\7\end{pmatrix}$A1
- Area $=\tfrac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\tfrac12\sqrt{2^2+3^2+7^2}$ using the magnitude of their cross productM1
- $\tfrac12\sqrt{62}=3.94$ (accept $\tfrac{\sqrt{62}}{2}$)A1
- Answer: (a) $\overrightarrow{AB}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$, $\overrightarrow{AC}=\begin{pmatrix}1\\4\\-2\end{pmatrix}$ (b) $\begin{pmatrix}2\\3\\7\end{pmatrix}$ (c) $\tfrac12\sqrt{62}\approx3.94$
- Examiner tip: Subtract in the order (end $-$ start) and remember the factor $\tfrac12$: the magnitude of the cross product is the area of the parallelogram.