Find the foot of the perpendicular from the point (2,3,-8) to the line `(4-x)/2 =y/6=(1-z)/3`. Find the perpendicular distance from the given point to the line.

Explanation

We have equation of line as `(4-x)/2=y/6=(1-z)/3`
`implies (x-4)/(-2)=y/6=(z-1)/(-3)=lamda`
`implies x=-2 lamda +4, y=6lamda` and `z=-3lamda+1`
Let the coordinates of `L` be `(4-2lamda, 6lamda,1-3lamda)` and direction ratios of `PL` are proportional to `(4-2lamda-2,6lamda-3,1-3lamda+8)` i.e., `(2-2lamda,6lamda-3,9-3lamda)`.
Also, direction ratios are proportional to `-2,5,-3`. Since `PL` is perpendicular to give line.
`:. -2(2-2lamda)+6(6lamda-3)-3(9-3lamda)=0`
`implies -4+4lamda+36lamda-18-27+9lamda=0`
`implies49lamda=49implieslamda=1`
So, the coordintes of `L` are `(4-2lamda,6lamda,1-3lamda)` i.e., `(2,6,-2)`

Also length of `PL=sqrt((2-2)^(2)+(6-3)^(2)+(-2+8)^(2))`
`=sqrt(0+9+36)=3sqrt(5)` units