## Calculus 8th Edition

$D=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$
$n=\lt a,b,c\gt$ $a(x_1)+b(0)+c(0)+d_1=0$ $P=(\frac{-d_1}{a},0,0)$ $a(x_2)+b(0)+c(0)+d_2=0$ $Q=(\frac{-d_2}{a},0,0)$ Take a vector $B$ for the line between points $P$ and $Q$. $B=PQ=(\frac{-d_1}{a},0,0) \cdot (\frac{-d_2}{a},0,0)$ $=(\frac{d_1-d_2}{a},0,0)$ $n.B=\lt d_1-d_2,0,0\gt$ $|n|=\sqrt {a^2+b^2+c^2}$ Therefore, the distance formula between two parallel planes is given as: $D=\frac{|n.B|}{|n|}=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$