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

Question

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

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Answer

The given equation of the plane is
x + 2y + 3z - 6 = 0
x + 2y + 3z = 6
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=6$
Or $\vec{\text{r}}\cdot\vec{\text{n}}=6$
When, $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ ...(\text{i})$
Now, $|\vec{\text{n}}|=\sqrt{1^2+2^2+3^2}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
Unit vector to the plane, $\hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{14}}$
$=\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}$

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