Question

Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]

 

Answer 1

\[ \text{ The required plane passes through the point } A(a, 0, 0) \text{ whose position vector is }  \vec{a} =a \hat{i} + 0 \hat{j}  + 0 \hat{k}  \text{ and is normal to the vector }  \vec{n}  \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC} . \]
\[ \text{ Clearly, } \vec{AB} = \vec{OB} - \vec{OA} = \left( 0 \hat{i} + b \hat{j} + 0 \hat{k}  \right) - \left( a \hat{i}  + 0 \hat{j}  + 0 \hat{k} \right) = - a \hat{i} + b \hat{j}  + 0 \hat{k}  \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( 0 \hat{i}  + 0 \hat{j}  + c \hat{k}  \right) - \left( a \hat{i}  + 0 \hat{j}  + 0 \hat{k}  \right) = - a \hat{i}  + 0 \hat{j}  + c \hat{k}  \]
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ - a & b & 0 \\ - a & 0 & c\end{vmatrix} =\text{bc }   \hat{i}  + \text{ ac }\hat{j}  + \text{ab } \hat{k}   \]
\[ \text{ The vector equation of the required plane is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( \text{ bc }\hat{i}  + \text{ac }  \hat{j}  + \text{ ab }  \hat{k}  \right) = \left( a \hat{i}  + 0  \hat{j}  + 0 \hat{ k} \right) . \left( \text{ bc }\hat{i}  +\text{ ac }\hat{j}  +\text{ ab }\hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left( \text{bc } \hat{i} + \text{ac }\hat{j} + \text{ab }\hat{k}  \right) = abc + 0 + 0\]
\[ \Rightarrow \vec{r} . \left( bc \hat{i} + ac \hat{j}  + ab \hat{k}  \right) = abc . . . \left( 1 \right)\]
\[\text{ Now } ,\left| \vec{n} \right|=\sqrt{\left( bc \right)^2 + \left( ac \right)^2 + \left( ab \right)^2}=\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}\]
\[ \text{ For reducing (1) to normal form, we need to divide both sides of (1) by } \sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}. \text{ Then, we get } \]
\[ \vec{r} . \left( \frac{bc \hat{i}  + ac \hat{j}  + ab \hat{k} }{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}} \right) = \frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}, \text{ which is the normal form of plane } (1).\]
\[ \text{ So, the distance of plane (1) from the origin,} \]
\[p = \frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}, \]
\[ \Rightarrow \frac{1}{p} = \frac{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}{abc}\]
\[ \Rightarrow \frac{1}{p^2} = \frac{b^2 c^2 + a^2 c^2 + a^2 b^2}{a^2 b^2 c^2}\]
\[ \Rightarrow \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\]
\[\]