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.Prove that the area of the triangle ABC is $\frac{r^5}{2fgh}$ where OP=r.
I think the direction normals of the plane should be (f,g,h)but I know how to proceed after that. Can someone suggest any approach?
I think there is an error...
Let $\Pi$ the plane. $P\in\Pi$ and $OP$ is a normal vector to $\Pi$. Thus, if $Q=(x,y,z)$ is in $\Pi$, we have $\vec{QP}\perp\vec{OP}$. But $\vec{QP}=\vec{OP}-\vec{OQ}$. Then, $0=\langle \vec{QP},\vec{OP}\rangle=\langle\vec{OP}-\vec{OQ},\vec{OP}\rangle=\langle\vec{OP},\vec{OP}\rangle-\langle\vec{OQ},\vec{OP}\rangle$, and since $\langle\vec{OP},\vec{OP}\rangle=r^2$, then $r^2=\langle\vec{OQ},\vec{OP}\rangle=fx+gy+hz$
So the equation of $\Pi$ is $fx+gy+hz-r^2=0$. Now, putting $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ in the last equation, we get the points of intersection of $\Pi$ with axes $X$, $Y$ and $Z$, respectively, giving us $A=(\frac{r^2}{f},0,0)$, $B=(0,\frac{r^2}{g},0)$ and $C=(0,0,\frac{r^2}{h})$.
Therefore, if $u=B-A$ and $v=C-A$, the area $A$ is given by $A=\dfrac{|u\times v|}{2}...(*)$
But $u=(-\frac{r^2}{f},\frac{r^2}{g},0)$ and $v=(-\frac{r^2}{f},0,\frac{r^2}{h})$. Therefore
$u\times v=\left|\begin{array}{ccc}i&j&k\\-\frac{r^2}{f}&\frac{r^2}{g}&0\\-\frac{r^2}{f}&0&\frac{r^2}{h}\end{array}\right|=\frac{r^4}{gh}i+\frac{r^4}{fh}j+\frac{r^4}{fg}k$, giving us
$|u\times v|=\sqrt{\dfrac{r^8}{g^2h^2}+\dfrac{r^8}{f^2h^2}+\dfrac{r^8}{f^2g^2}}=\sqrt{\dfrac{r^8}{f^2g^2h^2}(f^2+g^2+h^2)}=\sqrt{\dfrac{r^8}{f^2g^2h^2}r^2}=\dfrac{r^5}{fgh}$ (impossing $f,g,h>0$).
Finally, by $(*)$ we have $A=\dfrac{r^5}{2fgh}$