# Using vectors,find the area of triangle ABC,whose vertices are $A(1,2,3),B(2,-1,4),C(4,5,-1).$

Let $\overrightarrow{a},\overrightarrow{b}\; and \;\overrightarrow{c}$ be the position vectors of the triangle ABC
$\overrightarrow{a}=\overrightarrow{i}+2 \overrightarrow{j}+3 \overrightarrow{k}$
$\overrightarrow{b}=2 \overrightarrow{i}-\overrightarrow{j}+4 \overrightarrow{k}\;and\;\overrightarrow{c}=4 \overrightarrow{i}+5 \overrightarrow{j}-\overrightarrow{k}$
Area of $\bigtriangleup ABC=\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
$\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$
$=(2\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})-(\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})$
Therefore $\overrightarrow{AB}=\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}$
$=(4\overrightarrow{i}+5\overrightarrow{j}-\overrightarrow{k})-(\overrightarrow{i}+2 \overrightarrow{j}+3 \overrightarrow{k})$
$=3\overrightarrow{i}+3\overrightarrow{j}-4\overrightarrow{k}$
Therefore $area=\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
$=\frac{1}{2}|(\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}) \times (3\overrightarrow{i}+3 \overrightarrow{j}-4 \overrightarrow{k})|$
$\overrightarrow{AB} \times \overrightarrow{AC}=\begin {vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$
$=\overrightarrow{i}(12-3)-\overrightarrow{j}(-4-3)+\overrightarrow{k}(3+9)$
$=9\overrightarrow{i}+7\overrightarrow{j}+12\overrightarrow{k}$
$|\overrightarrow{AB} \times \overrightarrow{AC}|=\sqrt {9^2+7^2+12^2}=\sqrt {274}$
Therefore $area=\frac{1}{2} \sqrt {274}$
answered Mar 24, 2013 by