Question

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.

Answer 1

$$\begin{gathered} \text{The given equations are} \\ 2x-y+3z=4...\left(1\right) \\ \text{The second equation of the plane is} \\ 2x-y+3z=18...\left(2\right) \\ \text{We know that the distance between two planes}ax+by+cz=d_1\text{and}ax+by+cz=d_2\text{is}\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance is} \\ \frac{\left|18-4\right|}{\sqrt{2^2+{\left(-1\right)}^2+3^2}} \\ \text{=}\frac{\left|14\right|}{\sqrt{4+1+9}} \\ =\frac{14}{\sqrt{14}} \\ =\sqrt{14}\text{units} \end{gathered}$$