Question

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

 

Answer 1

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