Question

Through a point $P(f,g,h)$, a plane is drawn at right angles to $OP$ where '$O$' is the origin, to meet the coordinate axes in $A,B,C$. Then the area of the triangle $ABC$ is ........

• A. $\frac{r^5}{fgh}$ where $OP=r$
• B. $\frac{r^5}{2fgh}$ where $OP=r$
• C. $\frac{r^4}{2fgh}$ where $OP=r$
• D. $\frac{r^4}{fgh}$ where $OP=r$

Answer 1

The correct option is B $\frac{r^5}{2fgh}$ where $OP=r$

Here $OP=\sqrt{f^2+g^2+h^2}=r$
Therefore D.R's of $OP$ are
$\frac{f}{\sqrt{f^2+g^2+h^2}},\frac{g}{\sqrt{f^2+g^2+h^2}},\frac{h}{\sqrt{f^2+g^2+h^2}}$ or $\frac{f}{r},\frac{g}{r},\frac{h}{r}$
Since $OP$ is normal to the plane, therefore equation of plane is
$\frac{f}{r}x+\frac{g}{r}y+\frac{h}{r}z=r\Rightarrow fx+gy+hz=r^2$
$\therefore A=(\frac{r^2}{f},0,0),B=(0,\frac{r^2}{g},0),C=(0,0,\frac{r^2}{h})$
Now area of $△ABC$
$△=\sqrt{A_{xy}^2+A_{yz}^2+A_{zx}^2}$
where $A_{xy}^2$ is projecton of $△ABC$ on
$xy$ plane $-$ area of $△AOB$
Now $A_{xy}=\frac{1}{2}\left|\begin{array}{ccc}\frac{r^2}{f}& 0& 1\\ 0& \frac{r^2}{g}& 1\\ 0& 0& 1\end{array}\right|=\frac{r^4}{2\left|fg\right|}$
Similarly $A_{yz}=\frac{r^4}{2\left|gh\right|}$ and $A_{zx}=\frac{r^4}{2\left|hf\right|}$
$\therefore △=\sqrt{A_{xy}^2+A_{yz}^2+A_{zx}^2}=\frac{r^5}{2fgh}$