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

\[\text{ Let } \alpha, \beta \text{ and }  \gamma \text{ be the angles made by }\vec{n} \text{ withx, y andz-axes, respectively }.\]

\[\text{ It is given that } \]

\[\alpha = \beta = \gamma\]

\[ \Rightarrow \cos \alpha = \cos \beta = \cos \gamma\]

\[ \Rightarrow l = m = n, \text{ wherel, m, n are direction cosines of }  \vec{n} .\]

\[\text{ But } l^2 + m^2 + n^2 = 1\]

\[ \Rightarrow l^2 + l^2 + l^2 = 1\]

\[ \Rightarrow \text{ 3  l}^2 = 1\]

\[ \Rightarrow l^2 = \frac{1}{3}\]

\[ \Rightarrow l = \frac{1}{\sqrt{3}}\]

\[So,l = m = n = \frac{1}{\sqrt{3}}\]

\[\text{ It is given that the length of the perpendicular of the plane from the origin } ,p= 3\sqrt{3}\]

\[ \text{ The normal form of the plane islx + my + nz = p } \]

\[ \Rightarrow \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = 3\sqrt{3}\]

\[ \Rightarrow x + y + z = 3\sqrt{3} \left( \sqrt{3} \right) \]

\[ \Rightarrow x + y + z = 9\]