Find the length and the foot of the perpendicular from the point P(7,14,5) to the plane (2x+4y-z=2). Also, find the image of the point P in the plane.

Explanation

The equation of the given plane is
`2x+4y-z=2`……………(i)
The d.r's of the normal to the plane are 2,4,-1.
The equation of the line passing through the point P(7,14,5) and perpendicular to the given line (i) is given by

`(x-7)/2=(y-14)/4=(z-5)/-1=lambda`(say)
A general point on this line is `(2lambda+7,4lambda+14,-lambda+5)`.
For some value of `lambda`, let the point `Q(2lambda+7,4lambda+14,-lambda+5)` lie on the plane (i). Then,
`(2(2lambda+7)+4(4lambda+14)-(-lambda+5)=2 rArr 21lambda=-63 rArr lambda=-3`.
`therefore` the coordinates of the foot of the perpendicular PQ are
`Q[2 xx (-3)+7,4 xx (-3)+14,-(-3)+5]`, i.e., `Q(1,2,8)`.
`therefore PQ =sqrt(6^(2)+(12)^(2)+(-3)^(2))=sqrt(36+144+9)`
`=sqrt(189)=3sqrt(21)` units.
Let `R(x,y,z)` be the image of P in the plane (i).
Then, Q is the midpoint of PR.
`therefore (7+x)/2=1, (14+y)/2=2` and `(5+z)/2=8 rArr x=-5, y=-10, z=11`.
Hence, the iamge of `P(7,14,5)` in the given plane is `R(-5,-10,11)`.