Question

Find the equation of a plane which is at a distance $3\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axis.

Answer 1

Since, normal to the plane is equally inclined to the cooridinate axis.

Therefore,$cos\alpha =cos\beta =cos\gamma =\frac{1}{\sqrt{3}}$
So, the normal is $\overrightarrow{N}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$ and plane is at a distance of $3\sqrt{3}$ units from origin.
The equation of plane is $\overrightarrow{r}.\hat{N}=3\sqrt{3}[\because \hat{N}=\frac{\overrightarrow{N}}{|N|}]$
[since, vector equation of the plane at a a distance p from the origin is$\hat{r}.\hat{N}=p$]
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\frac{(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k})}{1}=3\sqrt{3}\Rightarrow \frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=3\sqrt{3}\therefore x+y+z=3\sqrt{3}.\sqrt{3}=9$.
So, the required equation of plane is x+y+z=9.