Question

Distance(in $\text{units}$) between the parallel planes $2x-3y+4z-1=0$ and $4x-6y+8z+7=0$ is

• A. $\frac{9}{\sqrt{13}}$
• B. $\frac{3}{\sqrt{29}}$
• C. $\frac{9}{2\sqrt{29}}$
• D. $\frac{7}{2\sqrt{13}}$

Answer 1

The correct option is C $\frac{9}{2\sqrt{29}}$

Given: $2x-3y+4z=1\Rightarrow 4x-6y+8z-2=0\cdots \left(i\right)$ and$,4x-6y+8z+7=0\cdots \left(ii\right)$
Comparing the given equations with $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$​ we get,
$a=4,b=-6,c=8,d_1=-2,d_2=7$
Distance between plane $1$ and $2$ $=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$
$=\frac{|-2-7|}{\sqrt{16+36+64}}=\frac{9}{\sqrt{116}}$
Hence, distance between the planes is $\frac{9}{2\sqrt{29}}$