Question

Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector $\hat{i}-2\hat{j}-2\hat{k}.$

Answer 1

$$\begin{gathered} \text{It is given that the normal vector,}\overrightarrow{n}\text{=}\hat{i}-2\hat{j}-2\hat{k} \\ \text{Now,}\hat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{1+4+4}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \\ \text{The equation of a plane in normal form is} \\ \overrightarrow{r}.\hat{n}=d\text{(where}\mathit{\text{d}}\text{is the distance of the plane from the origin)} \\ \text{Substituting}\hat{n}\text{=}\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}\text{and}\mathit{\text{d}}\text{= 5} \\ \mathrm{Here}, \\ \overrightarrow{r}.\left(\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}\right)=5 \end{gathered}$$