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 plane is 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,-5/3)`

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

We have
`PM=5`
`implies(|1-4-2-alpha|)/(sqrt(1+4+4))=5`
`implies|alpha+5|=15`
`implies alpha+5=+-15`
`impliesalpha=10,-20`
`impliesalpha=10 [ :' alpha gt 0]`
The equation of PM is
`(x-1)/1=(y+2)/2=(z-1)/(-2)`

The coordinates of `M` are given by
`(x-1)/1=(y+2)/2=(z-1)/(-2)=r`
So, the coordinates of `M` are `(r+1,2r-2,-2r+1)`. As `M` lies on the plane `x+2y-2z=alpha`.
`:.r+1+4r-4+4r-2=10 [ :' alpha=10]`
`implies 9r=15impliesr=5/3`
Hence the coordinates of `M` are `(8/3,4/3,-7/3)`