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

$$\begin{gathered} \text{The required plane passes through the point}\mathit{\text{A}}\text{(}\mathit{\text{a}}\text{, 0, 0) whose position vector is}\overrightarrow{a}\text{=}\mathit{\text{a}}\hat{i}\text{+ 0}\hat{j}\text{+ 0}\hat{k}\text{and is normal to the vector}\overrightarrow{n}\text{given by} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}. \\ \text{Clearly,}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(0\hat{i}+b\hat{j}+0\hat{k}\right)-\left(\mathit{\text{a}}\hat{i}\text{+ 0}\hat{j}\text{+ 0}\hat{k}\right)=-a\hat{i}+b\hat{j}+0\hat{k} \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(0\hat{i}+0\hat{j}+c\hat{k}\right)-\left(\mathit{\text{a}}\hat{i}\text{+ 0}\hat{j}\text{+ 0}\hat{k}\right)=-a\hat{i}+0\hat{j}+c\hat{k} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -a& b& 0\\ -a& 0& c\end{array}\right|=bc\hat{i}+ac\hat{j}+ab\hat{k} \\ \text{The vector equation of the required plane is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \Rightarrow \overrightarrow{r}.\left(bc\hat{i}+ac\hat{j}+ab\hat{k}\right)=\left(\mathit{\text{a}}\hat{i}\text{+ 0}\hat{j}\text{+ 0}\hat{k}\right).\left(bc\hat{i}+ac\hat{j}+ab\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(bc\hat{i}+ac\hat{j}+ab\hat{k}\right)=abc+0+0 \\ \Rightarrow \overrightarrow{r}.\left(bc\hat{i}+ac\hat{j}+ab\hat{k}\right)=abc...\left(1\right) \\ \text{Now,}\left|\overrightarrow{n}\right|\text{=}\sqrt{{\left(bc\right)}^2+{\left(ac\right)}^2+{\left(ab\right)}^2}\text{=}\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} \\ \overrightarrow{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} \end{gathered}$$