Determine the mutual position of the planes vec r·(vec i-vec…

Determine the mutual position of the planes $\vec r\cdot(\vec i-\vec j+\vec k)=3$ and $\vec r\cdot(\vec i-\vec j+\vec k)=-4$.

Explanation of Method

Two planes with the same normal vector are parallel; here they are at distance 7/√3.

The two planes have the same normal $\{1,-1,1\}$ and different right-hand sides, so they are parallel and distinct:

\[ d=\frac{|3-(-4)|}{|\vec n|}=\frac{7}{\sqrt3}=\frac{7\sqrt3}{3}. \]

\[\pi_1\parallel\pi_2,\qquad d=\frac{7\sqrt3}{3}\approx4.04\]

The origin lies between them, since the signs of $3$ and $-4$ differ. Its distances to the planes are $\tfrac3{\sqrt3}=\sqrt3$ and $\tfrac4{\sqrt3}$.

$O$ $d=\tfrac{7}{\sqrt3}$

Two parallel planes with common normal $\{1,-1,1\}$; the dashed segment along the normal (through $O$) has length $\tfrac7{\sqrt3}$.

\[\boxed{\,\pi_1\parallel\pi_2,\qquad d=\frac{7\sqrt3}{3}\approx4.04\,}\]