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In R^(3) let L be straight line passing ...

In `R^(3)` let L be straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes `P_(1):x+2y-z+1=0 and P_(2) : 2x-y+z-1=0` Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane `P_(1)` . which of the following points lie (s) on M?

A

`(0,-(5)/(6),-(2)/(3))`

B

`(-(1)/(6),-(1)/(3),(1)/(6))`

C

`(-(5)/(6),0,(1)/(6))`

D

`(-(1)/(3),0,(2)/(3))`

Text Solution

Verified by Experts

The correct Answer is:
A, B
Since, L is at constant distance from two planes `P_(1)` and `P_(2)`
Therefore, L is parallel to the line through intersection of `P_(1)` and `P_(2).`
DR's of `L=|{:(hati,hatj,hatk),(1,2,-1),(2,-1,1):}|`
`=hati(2-1)-hatj(1+2)+hatk(-1-4)`
`=hati-3hatj-5hatk`
`:.`DR's of L are (1,-3,-5) passing through (0,0,0).
Now, equation of L is
`(x-0)/(1)=(y-0)/(-3)=(z-0)/(-5)`
For any point on L, `(x)/(1)=(y)/(-3)=(z)/(-5)=lambda" "["say"]`
i.e.`" "P(lambda,-3lambda,-5lambda)`
If `(alpha,beta,gamma)` is foot of perpendicualr from P on `P_(1),` then
`(alpha-lambda)/(1)=(beta+3lambda)/(2)=(gamma+5lambda)/(-1)=k" "["say"]`
`impliesalpha=lambda+k,beta=2k-3lambda,gamma=-k-5lambda`
which satisfy `P_(1):x+2y-z+1=0`
`implies(lambda+k)+2(2k-3lambda)-(-k-5lambda)+1=0`
`implies" "k=-(1)/(6)`
`:." "x=-(1)/(6)+lambda,y=-(1)/(3)-3lambda,x=(1)/(6)-5lambda`
which satisfy options (a) and (b).
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