Determine the mutual position of the planes x-2y+3z=2 and…

Determine the mutual position of the planes $x-2y+3z=2$ and $2x-4y+6z=2$.

Explanation of Method

Two planes with proportional normals but non-proportional free terms are parallel and distinct.

Step 1: Normal vectors

\[ \vec n_1=\{1,-2,3\},\qquad \vec n_2=\{2,-4,6\}=2\vec n_1 . \]

The normals are parallel, so the planes are parallel or coincident.

Step 2: Coincident or not?

Dividing the second equation by $2$ gives $x-2y+3z=1$. The left-hand sides now agree, but $2\ne1$. Equivalently, the ratios $\tfrac12=\tfrac{-2}{-4}=\tfrac36\ne\tfrac22$ are not all equal. So there is no common point.

Step 3: Distance

\[ d=\frac{|2-1|}{\sqrt{1+4+9}}=\frac{1}{\sqrt{14}} . \]

\[\text{parallel and distinct},\qquad d=\frac{\sqrt{14}}{14}\approx0.267\]

Check: the point $(2,0,0)$ of the first plane has distance $\tfrac{|2-1|}{\sqrt{14}}$ from $x-2y+3z-1=0$.

$O$

$x-2y+3z=2$
$x-2y+3z=1$

Two parallel planes with common normal $\{1,-2,3\}$. Along the normal through $O$ they lie at distances $\tfrac{1}{\sqrt{14}}$ and $\tfrac{2}{\sqrt{14}}$ from $O$, on the same side, so $d=\tfrac{1}{\sqrt{14}}$.

Result

\[\boxed{\,\text{parallel and distinct},\qquad d=\frac{\sqrt{14}}{14}\approx0.267\,}\]