Question

Through a point $P(h,k,l)$ a plane is drawn at right angles to $OP$ to meet the coordinate axes in $A,B$ and $C$. If $OP=p,$ then the area of $△ABC$ is

• A. $\frac{p^5}{2hkl}$
• B. $\frac{p^5}{hkl}$
• C. $\frac{p^5}{4hkl}$
• D. None of these

Answer 1

The correct option is A $\frac{p^5}{2hkl}$

Here, $OP=\sqrt{h^2+k^2+l^2}=p$

$\therefore$ Dr's of $OP$ are: $\frac{h}{\sqrt{h^2+k^2+l^2},}\frac{k}{\sqrt{h^2+k^2+l^2}},\frac{l}{\sqrt{h^2+k^2+l^2}}$ or $\frac{h}{p},\frac{k}{p},\frac{l}{p}$

Since $OP$ is normal to plane, therefore, equation of plane

$\frac{h}{p}x+\frac{k}{p}y+\frac{l}{p}z=phx+ky+lz=p^2$

$A\equiv (\frac{p^2}{h},0,0),B\equiv (0,\frac{p^2}{k},0),C\equiv (0,0,\frac{p^2}{l})$

Now, Area of $△ABC,△=\sqrt{A_{xy}^2+A_{yx}^2+A_{zx}^2}$

where, $A_{xy}^2$ is area of projection of $△ABC$ on $xy$ plane $=$ area of $△AOB$

Now, $A_{xy}=\frac{1}{2}\left|\begin{array}{ccc}\frac{p^2}{h}& o& 1\\ 0& \frac{p^2}{k}& 1\\ 0& 0& 1\end{array}\right|=\frac{p^4}{2\left|hk\right|}$

Similarly, $A_{yz}=\frac{p^4}{2\left|kl\right|}$ and $A_{zx}=\frac{p^4}{2\left|lh\right|}$

$\therefore △=\sqrt{A_{xy}^2+A_{yz}^2+A_{zx}^2}=\frac{p^5}{2hkl}$