Question

Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

Answer 1

$$\begin{gathered} \text{The given equation of the plane is} \\ 2x-3y+6z+14=0 \\ 2x-3y+6z=-14...\left(1\right) \\ \text{Now,}\sqrt{2^2+{\left(-3\right)}^2+{\left(6\right)}^2}\text{=}\sqrt{4+9+36}\text{=}\sqrt{49}\text{= 7} \\ \text{Dividing (1) by 7, we get} \\ \frac{2}{7}x-\frac{3}{7}y+\frac{6}{7}z=-2 \\ \text{Multiplying both sides by -1, we get} \\ -\frac{2}{7}x+\frac{3}{7}y-\frac{6}{7}z=2 \\ \text{This is the normal form of the given equation of the plane.} \end{gathered}$$