Find the distance between the planes: 1^° x-2y+z-1=0 and…

Find the distance between the planes: $1^\circ$ $x-2y+z-1=0$ and $2x-4y+2z+1=0$; $2^\circ$ $\vec r\cdot(2\vec i+3\vec j-6\vec k)+14=0$ and $\vec r\cdot(2\vec i+3\vec j-6\vec k)-35=0$.

Step 1: Formula

For $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ (with the same coefficients): $d=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$.

Step 2: $1^\circ$

$\vec n_2=2\vec n_1$, so the planes are parallel. Halving the second equation gives $x-2y+z+\tfrac12=0$: $d=\frac{\left|-1-\tfrac12\right|}{\sqrt6}=\frac{3}{2\sqrt6}$.

\[d=\frac{\sqrt6}{4}\]

Check: the point $P(1,0,0)$ of the first plane has distance $\tfrac{|2+1|}{\sqrt{24}}=\tfrac{\sqrt6}{4}$ from the second.

Step 3: $2^\circ$

The common normal is $\{2,3,-6\}$, with $|\vec n|=7$: $d=\frac{|14-(-35)|}{7}=7$.

\[d=7\]

The origin lies between the planes, at distances $2$ and $5$.