Question

The distance between the planes 2x + 2y − z + 2 = 0 and 4x + 4y − 2z + 5 = 0 is
(a) $\frac{1}{2}$

(b) $\frac{1}{4}$

(c) $\frac{1}{6}$

(d) None of these

Answer 1

(c) $\frac{1}{6}$

$$\begin{gathered} \text{Multiplying the first equation of the plane by} \\ 4x+4y-2z+4=0 \\ 4x+4y-2z=-4...\left(1\right) \\ \text{The second equation of the plane is} \\ 4x+4y-2z+5=0 \\ 4x+4y-2z=-5...\left(2\right) \\ \text{We know that the distance between two planes}ax+by+cz=d_1\text{and}ax+by+cz=d_2\text{is}\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance} \\ \text{=}\frac{\left|-5+4\right|}{\sqrt{4^2+4^2+{\left(-2\right)}^2}} \\ \text{=}\frac{\left|-1\right|}{\sqrt{16+16+4}} \\ =\frac{1}{\sqrt{36}} \\ =\frac{1}{6}\text{units} \end{gathered}$$