Question

Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.

 

Answer 1

\[\text{ Let } \alpha, \beta \text{ and }  \gamma \text{ be the angles made by }  \vec{n} \text{ with x, y and z-axes, respectively } .\]
\[\text{ It is given that } \]
\[\alpha = \beta = \gamma\]
\[ \Rightarrow \cos\alpha = \cos\beta = \cos\gamma\]
\[ \Rightarrow l = m = n, \text{ where l, m, n are direction cosines of }  \vec{n} .\]
\[\text{ But } l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow l^2 + l^2 + l^2 = 1\]
\[ \Rightarrow 3 l^2 = 1\]
\[ \Rightarrow l^2 = \frac{1}{3}\]
\[ \Rightarrow l = \frac{1}{\sqrt{3}}\]
\[\text{ So } ,l = m = n = \frac{1}{\sqrt{3}}\]
\[\text{ It is given that the length of the perpendicular of the plane from the origin, } p= 5\sqrt{3}\]
\[\text{ The normal form of the plane is } lx + my + nz = p\]
\[ \Rightarrow \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = 5\sqrt{3}\]
\[ \Rightarrow x + y + z = 5\sqrt{3} \left( \sqrt{3} \right) \]
\[ \Rightarrow x + y + z = 15\]