Question

Let any tangent plane to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$ makes intercepts $a,b,c$ with the coordinate axes at $A,B,C$ respectively. If $P$ is the centre of the sphere, then
$($$ar.$ and $vol.$ denote the area and volume respectively$)$

• A. $vol.\left(PABC\right)=\frac{abc}{3}$
• B. $ar.\left(△ABC\right)=\frac{abc}{r}$
• C. $ar.\left(△PAB\right)=\frac{abc}{r}$
• D. $vol.\left(PABC\right)=\frac{abc}{6}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $vol.\left(PABC\right)=\frac{abc}{3}$
B $ar.\left(△ABC\right)=\frac{abc}{r}$
C $ar.\left(△PAB\right)=\frac{abc}{r}$

Equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Distance of the plane from the centre $P(a,b,c)$ of the sphere is
$\frac{2}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=r\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{4}{r^2}$

Now, $\overrightarrow{AB}=-a\hat{i}+b\hat{j}$ and $\overrightarrow{AC}=-a\hat{i}+c\hat{k}$
$ar.\left(△ABC\right)=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -a& b& 0\\ -a& 0& c\end{array}\right|\right|=\frac{1}{2}\sqrt{(ab{)}^2+(bc{)}^2+(ca{)}^2}=\frac{abc}{2}\sqrt{\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}}=\frac{abc}{r}$

$\overrightarrow{PA}=-b\hat{j}-c\hat{k}$
$ar.\left(△PAB\right)=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{PA}|=\frac{1}{2}\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -a& b& 0\\ 0& -b& -c\end{array}\right|\right|=\frac{1}{2}\sqrt{(ab{)}^2+(bc{)}^2+(ca{)}^2}=\frac{abc}{2}\sqrt{\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}}=\frac{abc}{r}$

$vol.\left(PABC\right)=\frac{1}{3}\times \text{base area}\times \text{height}=ar.\left(△ABC\right)\times \frac{r}{3}=\frac{abc}{3}$