Question

A plane makes intercepts $OA,OB$ and $OC$ whose measurements are $a,b$ and $c$ on the $OX,OY$ and $OZ$ axes. The area of triangle $ABC$ is-

• A. $\frac{1}{2}(ab+bc+ca)$
• B. $\frac{1}{2}abc(a+b+c)$
• C. $\frac{1}{2}(a^2{b}^2+b^2{c}^2+c^2{a}^2{)}^{1/2}$
• D. $\frac{1}{2}(a+b+c{)}^2$

Answer 1

The correct option is C $\frac{1}{2}(a^2{b}^2+b^2{c}^2+c^2{a}^2{)}^{1/2}$

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, $A=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$