Question

Reduce the equation $\overrightarrow{r}·\left(\hat{i}-2\hat{j}+2\hat{k}\right)+6=0$ to normal form and, hence, find the length of the perpendicular from the origin to the plane.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ \overrightarrow{r}.\left(\hat{i}-2\hat{j}+2\hat{k}\right)+6=0 \\ \Rightarrow \overrightarrow{r}.\left(\hat{i}-2\hat{j}+2\hat{k}\right)=-6\text{or}\overrightarrow{r}.\overrightarrow{n}=-6,\text{where}\overrightarrow{n}\text{=}\hat{i}-2\hat{j}+2\hat{k} \\ \left|\overrightarrow{n}\right|=\sqrt{1+4+4}=3 \\ \text{For reducing the given equation to normal form, we need to divide it by}\left|\overrightarrow{n}\right|\text{. Then, we get} \\ \overrightarrow{r}.\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{-6}{\left|\overrightarrow{n}\right|} \\ \Rightarrow \overrightarrow{r}.\left(\frac{\hat{i}-2\hat{j}+2\hat{k}}{3}\right)=\frac{-6}{3} \\ \Rightarrow \overrightarrow{r}.\left(\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}\right)=-2 \\ \text{Dividing both sides by -1, we get} \\ \overrightarrow{r}.\left(-\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}\right)=2...\left(1\right) \\ \text{The equation of the plane in normal form is} \\ \overrightarrow{r}.\hat{n}=d...\left(2\right) \\ \text{(where}\mathit{\text{d}}\text{is the distance of the plane from the origin)} \\ \text{Comparing (1) and (2),} \\ \mathrm{l}\text{ength of the perpendicular from the origin to the plane =}\mathit{\text{d}}\text{= 2 units} \end{gathered}$$