Question

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

• A. $\frac{1}{2}\sum$ $ab$
• B. $\frac{1}{2}\left[\right(ab{)}^2+(bc{)}^2+(ca{)}^2{]}^{1/2}$
• C. $\frac{abc}{2}\sum a$
• D. $\frac{1}{2}[\sum a\sum ab]$

Answer 1

The correct option is B $\frac{1}{2}\left[\right(ab{)}^2+(bc{)}^2+(ca{)}^2{]}^{1/2}$

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