## Calculus: Early Transcendentals 8th Edition

$x+2y-2z-7=0$ ; $x+2y-2z+5=0$
$x+2y-2z=1$ $x+2y-2z-1=0$ $n= \lt 1,2, -2\gt$ Distance formula between two parallel planes is given as: $D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$ $2=\frac{|x+2y-2z-1|}{\sqrt {1^2+2^2+(-2)^2}}$ $6=|x+2y-2z-1|$ First plane is: $6=x+2y-2z-1$ $\implies$ $x+2y-2z-7=0$ Second plane is: $6=-(x+2y-2z-1)$ $\implies$ $x+2y-2z+5=0$ Therefore, the distance formula between two parallel planes is given as: $D=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|-1-(-7)|}{3}$ $D=\frac{6}{3}=2$