Question

Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ x+2y+3z-6=0 \\ x+2y+3z=6 \\ \Rightarrow \overrightarrow{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)=6\text{or}\overrightarrow{r}\text{.}\overrightarrow{n}\text{= 6,} \\ \text{where}\overrightarrow{n}\text{=}\hat{i}+2\hat{j}+3\hat{k}...\left(1\right) \\ \text{Now,}\left|\overrightarrow{n}\right|\text{=}\sqrt{1^2+2^2+3^2}\text{=}\sqrt{1+4+9}\text{=}\sqrt{14} \\ \text{Unit vector to the plane,}\stackrel{⏜}{n}\text{=}\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\text{=}\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k} \end{gathered}$$