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?

Options

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))`

Answer

a, b

Explanation

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.