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Q.
The area of the triangle having vertices $A(1, 0, 0), B(0, 1, 0), C (0, 0, 1)$ equals
Three Dimensional Geometry
Solution:
Area of $\Delta\,ABC$, $A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}), C(x_{3}, y_{3}, z_{3})$ is giv s by
$\Delta=\sqrt{\Delta_{x}^{2}+\Delta_{y}^{2}+\Delta_{z}^{2}}$
$\left|\Delta_{x}\right|=\left|\Delta_{y}\right|=\left|\Delta_{z}\right|=\frac{1}{2}$, where $\Delta_{x}=\frac{1}{2} \begin{vmatrix}y_{1}&y_{2}&y_{3}\\ z_{1}&z_{2}&z_{3}\\ 1&1&1\end{vmatrix}$
$\therefore \Delta=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$
$=\frac{1}{2}\sqrt{3} $ sq. units Short Cut Method :
Area of $\Delta\,ABC =\frac{1}{2} \left|\overrightarrow{AB} \times\overrightarrow{AC}\right|$
where $\overrightarrow{AB} =\overrightarrow{OB}-\overrightarrow{OA}=-\hat{i} +\hat{j} $
and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-\hat{i}+\hat{k}$
$\therefore $ Area of $\Delta ABC =\frac{1}{2} \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -1&1&0\\ -1&0&1\end{vmatrix}$
$=\frac{1}{2}\left|\hat{i}+\hat{j}+\hat{k}\right|$
$=\frac{\sqrt{3}}{2}$ square units
