Question

Find the distance between the parallel planes 2x − y + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0.

Answer 1

$$\begin{gathered} \text{Multiplying the first equation of the plane by 3, we get} \\ 6x-3y+9z-12=0 \\ 6x-3y+9z=12...\left(1\right) \\ \text{The second equation of the plane is} \\ 6x-3y+9z=-13...\left(2\right) \\ \text{We know that the distance between two planes}ax+by+cz=d_1\text{and}ax+by+cz=d_2\text{is}\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance} \\ \text{=}\frac{\left|-13-12\right|}{\sqrt{6^2+{\left(-3\right)}^2+9^2}} \\ \text{=}\frac{\left|-25\right|}{\sqrt{36+9+81}} \\ =\frac{25}{\sqrt{126}} \\ =\frac{5}{3\sqrt{14}}\text{units} \end{gathered}$$