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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.
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Solution
\[\text{ Let } \alpha, \beta \text{ and } \gamma \text{ be the angles made by }\vec{n} \text{ withx, y andz-axes, respectively }.\]
\[\text{ It is given that } \]
\[\alpha = \beta = \gamma\]
\[ \Rightarrow \cos \alpha = \cos \beta = \cos \gamma\]
\[ \Rightarrow l = m = n, \text{ wherel, m, n are direction cosines of } \vec{n} .\]
\[\text{ But } l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow l^2 + l^2 + l^2 = 1\]
\[ \Rightarrow \text{ 3 l}^2 = 1\]
\[ \Rightarrow l^2 = \frac{1}{3}\]
\[ \Rightarrow l = \frac{1}{\sqrt{3}}\]
\[So,l = m = n = \frac{1}{\sqrt{3}}\]
\[\text{ It is given that the length of the perpendicular of the plane from the origin } ,p= 3\sqrt{3}\]
\[ \text{ The normal form of the plane islx + my + nz = p } \]
\[ \Rightarrow \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = 3\sqrt{3}\]
\[ \Rightarrow x + y + z = 3\sqrt{3} \left( \sqrt{3} \right) \]
\[ \Rightarrow x + y + z = 9\]
