Question

The area of the triangle having vertices $A(1,0,0),B(0,1,0),C(0,0,1)$ equals

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

Answer: (A)

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 ${\Delta }_x=\frac{1}{2}\left|\begin{array}{ccc}{y}_1& y_2& y_3\\ z_1& z_2& z_3\\ 1& 1& 1\end{array}\right|$
$\therefore \Delta =\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$
$=\frac{1}{2}\sqrt{3}$ 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 $\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& 0\\ -1& 0& 1\end{array}\right|$

$=\frac{1}{2}\left|\hat{i}+\hat{j}+\hat{k}\right|$
$=\frac{\sqrt{3}}{2}$ square units