Question

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 

Answer 1

\[ \text{ The given equation of the plane is } \]
\[2x - 3y - 6z = 14 . . . \left( 1 \right)\]
\[\text{ Now },\sqrt{2^2 + \left( - 3 \right)^2 + \left( - 6 \right)^2}=\sqrt{4 + 9 + 36}=\sqrt{49}= 7\]
\[\text{ Dividing (1) by 7, we get } \]
\[\frac{2}{7}x - \frac{3}{7}y - \frac{6}{7}z = 2 . . . \left( 2 \right)\]
\[\text{ The Cartesian equation of the normal form of a plane is } \]
\[lx + my + nz = p . . . \left( 3 \right), \]
\[ \text{ where l, m and n are direction cosines of normal to the plane and p is the length of the perpendicular from the origin to the plane }.\]
\[ \text{ Comparing (1) and (2), we get } \]
\[ \text{ direction cosines } : l = \frac{2}{7}, m = \frac{- 3}{7}, n = \frac{- 6}{7} \text{ and } \]
\[ \text{ length of the perpendicular from the origin to the plane } : p = 2\]