Question

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

Answer 1

$$\begin{gathered} \text{Let α, β and γ be the angles made by}\overrightarrow{n}\text{with}x,y\text{and}\mathit{\text{z}}\text{-axes, respectively.} \\ \text{It is}\text{given that} \\ \alpha =\beta =\gamma \\ \Rightarrow \cos \alpha =\cos \beta =\cos \gamma \\ \Rightarrow l=m=n,\text{where}l,m,n\text{are direction cosines of}\overrightarrow{n}\text{.} \\ \text{But}{l}^2+m^2+n^2=1 \\ \Rightarrow l^2+l^2+l^2=1 \\ \Rightarrow 3l^2=1 \\ \Rightarrow l^2=\frac{1}{3} \\ \Rightarrow l=\frac{1}{\sqrt{3}} \\ \text{So,}l=m=n=\frac{1}{\sqrt{3}} \\ \text{It is given that the length of the perpendicular of the plane from the origin,}\mathit{\text{p}}\text{= 3}\sqrt{3} \\ \text{The normal form of the plane is}lx+my+nz=p \\ \Rightarrow \frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=3\sqrt{3} \\ \Rightarrow x+y+z=3\sqrt{3}\left(\sqrt{3}\right) \\ \Rightarrow x+y+z=9 \end{gathered}$$