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

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• Area of a triangle =$\large\frac{1}{2}$$| \overrightarrow{AB} \times \overrightarrow{BC}| Given \overrightarrow {OA}=\hat i+2\hat j+3\hat k, \overrightarrow {OB}=2\hat i-\hat j+4\hat k, and\overrightarrow {OC}=4\hat i+5\hat j-\hat k, Area of a triangle is given by = \large\frac{1}{2}$$ | \overrightarrow{AB} \times \overrightarrow{BC}|$
Now let us determine $\overrightarrow{AB}$
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad=(2\hat i-\hat j+4\hat k)-(\hat i+2\hat j+3\hat k)$
$\qquad=\hat i-3\hat j+\hat k$
Now let us determine $\overrightarrow{AC}$
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
$\qquad=(4\hat i+5\hat j - \hat k)-(\hat i+2\hat j+3\hat k)$
$\qquad=3\hat i+3\hat j-4\hat k$
Now let us determine $\overrightarrow{AB} \times \overrightarrow{AC}$
$\overrightarrow{AB} \times \overrightarrow{AC}=\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$
$\qquad\qquad =\hat i(12-3)-\hat j(-4-3)+\hat k(3+9)$
$\qquad\qquad=9\hat i+7\hat j+12\hat k$
$| \overrightarrow{AB} \times \overrightarrow{AC}| =\sqrt{9^2+7^2+12^2}$
$\quad\qquad\qquad=\sqrt{81+49+144}$
$\quad\qquad\qquad=\sqrt{274}$
Hence the area of the triangle is $\large \frac{\sqrt{274}}{2}$ sq. units