Question

Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.

Answer 1

$$\begin{gathered} \text{Let the equation of a plane parallel to the given plane be} \\ 2x-3y+5z=k...\left(1\right) \\ \text{This passes through (3, 4, -1).\hspace{0.17em} So,} \\ 2\left(3\right)-3\left(4\right)+5\left(-1\right)=k \\ \Rightarrow k=-11 \\ \text{Substituting this in (1), we get} \\ 2x-3y+5z=-11...\left(1\right),\text{which is the equation of the required plane.} \\ \text{The equation of the given plane is} \\ 2x-3y+5z=-7...\left(2\right) \\ \text{We know that the distance between two planes}ax+by+cz=d_1\text{and}ax+by+cz=d_2\text{is}\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance} \\ =\frac{\left|-7-\left(-11\right)\right|}{\sqrt{2^2+{\left(-3\right)}^2+5^2}} \\ =\frac{\left|-7+11\right|}{\sqrt{4+9+25}} \\ =\frac{4}{\sqrt{38}}\text{units} \end{gathered}$$