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let L be a 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))`

A

`(0, - (5)/(9), - (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
Let `vecv` be the vector along `L`.
Clearly, line `L` is parallel to given planes
`P_(1) :x+2y-z+1=0 and P_(2) : 2x-y+z-1=0`
`therefore" "vecv= |{:(hati,,hatj,,hatk),(1,,2,,-1),(2,,-1,,1):}|= hati-3hatj-5hatk`
Since L is passing through origin, any point on line L is `A(lamda, -3lamda, -5lamda)`.
Foot of perpendicular from the A to `P_(1)`, is
`(h-lamda)/(1)= (k+3lamda)/(2)= (l+5lamda)/(-1)= ((lamda-6lamda+5lamda+1))/(1+4+1)= - (1)/(6)`
`therefore " "h=lamda- (1)/(6) , k= -3lamda- (1)/(3), l= -5lamda+ (1)/(6)`
So, foot or point on locus M is
`(lamda- (1)/(6), -3lamda- (1)/(3), -5lamda+ (1)/(6))`.
So points (a) and (b) lies on this locus.
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