If the distance of the point `P(1,-2,1)` from the plane `x+2y-2z=alpha,w h e r ealpha>0,i s5,` then the foot of the perpendicular from `P` to the place is

Options

A. `((8)/(3),(4)/(3)-(7)/(3))`
B. `((4)/(3),-(4)/(3),(1)/(3))`
C. `((1)/(3),(2)/(3),(10)/(3))`
D. `((2)/(3),-(1)/(3),(5)/(2))`

Answer

A

Explanation

Distance of point P from plane = 5
`:." "5=|{:(1-4-2-alpha)/(3):}|`
`rArr" "|5+alpha|=15`
`rArr" "alpha=10" "[becausealpha gt0]`
Foot of perpendicular, from P (1,-2,1) is
`(x-1)/(1)=(y+2)/(2)=(z-1)/(-2)=(-(1-4-2-10))/(1+4+4)=(5)/(3)`
[`because` foot (x,y,z) of a point `(x_(1),y_(1),z_(1))` in a plane ax+by+cz+d=0 is given by `(x-x_(1))/(a)=(y-y_(1))/(b)=(z-z_(1))/(c)=-((ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))]`
`rArr" "x=(8)/(3),y=(4)/(3),z=-(7)/(3)`
Thus, the foot of the perpendicular is
`A((8)/(3),(4)/(3),-(7)/(3))`