Question

From the distance of the point A($-1,-5,-10$) from a plane $x-y+z=5$ measured parallel to a line : $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$

Answer 1

$A(-1,-5,-10)$

$x-y+z=5$

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$

Required line equation is:

$\frac{x-2+1}{3}=\frac{y+1+5}{4}=\frac{z-2+10}{12}$

$\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+8}{12}=k$

$x=3k+1,y=4k-6,z=12k-8$

Given, $x-y+z=5$

$3k+1-4k+6+12k-8=5$

$11k=6$

$k=\frac{6}{11}$

$x=3k+1=3\left(\frac{6}{11}\right)+1=\frac{29}{11}$

$y=4k-6=4\left(\frac{6}{11}\right)-6=\frac{-42}{11}$

z=12k-8$=12\left(\frac{6}{11}\right)-8=\frac{-16}{11}$

distance between $(\frac{29}{11},\frac{-42}{11},\frac{-16}{11}),(-1,-5,-10)$

$=\sqrt{{(-1-\frac{29}{11})}^2+{(-5+\frac{42}{11})}^2+{(-10+\frac{16}{11})}^2}$

$=\frac{\sqrt{10605}}{11}$