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 $OP=p,A_{xy}$ is area of projection of $\Delta$ $ABC$ on $xy$-plane, $A_{xy}$ is area of projection of $\Delta$ABC on $yz$-plane, then

• A. $\Delta =\left|\frac{p^5}{2hkl}\right|$
• B. $\Delta =\left|\frac{p^5}{hkl}\right|$
• C. $\frac{A_{xy}}{A_{yz}}=\left|\frac{h}{l}\right|$
• D. $\frac{A_{xy}}{A_{yz}}=\left|\frac{l}{h}\right|$

Answer 1

Answer: (A), (D)

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

Therefore DRs of $OP$ are

$\frac{h}{\sqrt{h^2+k^2+l^2}},\frac{k}{\sqrt{h^2+k^2+l^2}},\frac{1}{\sqrt{h^2+k^2+l^2}}$ or $\frac{h}{p},\frac{k}{p},\frac{l}{p}$

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

$\frac{h}{p}x+\frac{h}{p}y+\frac{l}{p}z=p$

$\Rightarrow hx+ky+lz=p^2$

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

Now area of $△ABC$

$△=\sqrt{A_{xy}^2+A_{yz}^2+A_{zx}^2}$

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

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

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

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