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

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