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If the distance of the point P(1,-2,1...

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 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)`

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))`

Text Solution

Verified by Experts

The correct Answer is:
A
Distance of point P from plane =5
`:." "5=|(1-4-2-alpha)/(3)|`
`implies" "alpha=10`

Foot of perpendicular
`(x-1)/(1)=(y+2)/(2)=(z-1)/(-2)=(5)/(3)`
`implies" "x=(8)/(3),y=(4)/(3),z=-(7)/(3)`
Thus, the foot of the perpendicular is `A((8)/(3),(4)/(3),-(7)/(3)).`
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