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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 `(1, -2, 1)` from plane `x+2y-2z=alpha ` is `5rArr alpha = 10`.
Equation of `PQ, (x-1)/(1)= (y+2)/(2)=(z-1)/(-2)=t`
`Q-=(t+1, 2t-2, -2t+1) and PQ=5`
`rArr" "t= (5+alpha)/(9)= (5)/(3) rArr Q-= ((8)/(3), (4)/(3), (-7)/(3))`
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