Find the distance from M_0(2,0,1) to the planes: a)…

Find the distance from $M_0(2,0,1)$ to the planes: a) $x-2y+3z+1=0$; b) $\vec r\cdot(-\vec i+\vec j-\vec k)=2$.

Explanation of Method

The distance from the point M₀(2, 0, 1) to two planes.

\[ d=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} \]

\[ d=\frac{|2-0+3+1|}{\sqrt{14}}=\frac{6}{\sqrt{14}} \]

\[d=\frac{3\sqrt{14}}{7}\approx1.604\]

The foot of the perpendicular is $N=M_0-\tfrac{6}{14}\{1,-2,3\}=\left(\tfrac{11}7,\tfrac67,-\tfrac27\right)$.

The scalar form is $-x+y-z-2=0$: $d=\frac{|-2+0-1-2|}{\sqrt3}=\frac{5}{\sqrt3}$.

\[d=\frac{5\sqrt3}{3}\approx2.887\]

$M_0$ $N$ $d=\tfrac{3\sqrt{14}}{7}$

a) The perpendicular from $M_0(2,0,1)$ to $x-2y+3z+1=0$ meets it at $N\left(\tfrac{11}7,\tfrac67,-\tfrac27\right)$; $|M_0N|=\tfrac{6}{\sqrt{14}}$.

\[\boxed{\,d=\frac{3\sqrt{14}}{7}\approx1.604\,}\]

\[\boxed{\,d=\frac{5\sqrt3}{3}\approx2.887\,}\]