Find the co-ordinates of the foot of perpendicular and its perpendicular distance drawn from the point `(1,3,4)` to the plane `2x-y+2+3=0.`Also find the image of the point `(1,3,4)` in the plane.

Explanation

Let PN be the perpendicular from point `P (1,2,4)`to the plane `2x+y-2z+3 = 0`
`:.` Equation of PN
`(x-1)/(2) = (y-2)/(1) = (z-4)/(-2) = lambda`
(Say)
Let the variable point on this line be `(2lambda + 1, lambda + 2, -2lambda+4)`. This point lies on the plane `2x+y+2z+3 = 0`
`:. 2(2lambda+1)+(lambda+2)-2(-2lambda+4)+3=0`
`rArr 9 lambda - 1 = 0`
`rArr lambda = 1/9`
`:.` Co-ordinates of point N
`= (2/9+1,1/9+2,(-2)/(9)+4) = (11/9,19/9,34/9)`
and `PN = sqrt((11/9-1)^(2)+(19/9-2)^(2)+(34/9-4)^(2))`
`= sqrt(4/81+1/81+4/81)=1/3`
`:.` Required equation of perpendicular
`(x-1)/(2) = (y-2)/(1) = (z-4)/(-2)`
Length of perpendicular `= 1/3` units.
Front of perpendicular `= (11/9,19/9,34/9)`