Q. A plane makes intercepts $OA,OB$ and $OC$ whose measurements are $b$ 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) $\frac{1}{2}\left( a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} \right)^{1/2}$
D) $\frac{1}{2}(a + b + c)^{2}$
Explanation
(c)
Plane meets axes at $A(a,\ 0,0),\ B(0,\ b,\ 0)$ and $C(0,\ 0,\ c)$
Then area of $\Delta ABC$,
$= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
$= \frac{1}{2}|( - a\widehat{i} + b\widehat{j}) \times ( - a\widehat{i} + c\widehat{k})|$
$= \frac{1}{2}\sqrt{(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.