Question

If a plane makes intercepts of length $OA=a,OB=b,OC=c$ on axes of $OX,OY,OZ$, then area of $\Delta .ABC$ is given by

• A. $\frac{1}{2}\sqrt{a^2+b^2+c^2}$,
• B. $\frac{1}{2}ab\sin C$
• C. $\frac{1}{2}\sqrt{{\left(ab\right)}^2+{\left(bc\right)}^2+{\left(ca\right)}^2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}\sqrt{{\left(ab\right)}^2+{\left(bc\right)}^2+{\left(ca\right)}^2}$

Required area $=\sqrt{{\Delta }^{2}xy+{\Delta }^{2}yz+{\Delta }^{2}zx}=A$ (say)
where
$\Delta xy=\frac{1}{2}\mid \left|\begin{array}{ccc}a& 0& 1\\ 0& b& 1\\ 0& 0& 1\end{array}\right|\mid =\frac{1}{2}ab$

$\Delta yz=\frac{1}{2}\mid \left|\begin{array}{ccc}0& 0& 1\\ b& 0& 1\\ 0& c& 1\end{array}\right|\mid =\frac{1}{2}bc$

and $\Delta yx=\frac{1}{2}\mid \left|\begin{array}{ccc}0& c& 1\\ a& 0& 1\\ 0& 0& 1\end{array}\right|\mid =\frac{1}{2}ac$
Hence, $A=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
Also $A=\frac{1}{2}.|c-b|.|a-c|.\sin C$