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))`

Options

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))`

Answer

A, B

Explanation

a.b .
Let `vecv` be the vetor along I.
`P_(1):x+2y-Z+1=0 and P_(2):2x-y+z-1=0`
` therefore vecc=|{:(hati,hatj,hatk),( 1,2,-1),( 2,-1,1)}|= hati-3hatj-5hatk`
since I. is through origin any point on line I. is `A(lamda,3lamda ,-5lamda).`Foot of perpendicular from A to `P_(1)` , is
`(h- lamda)/( 1)`=(k+3lamda)/(-1)=-(lamda-6lamda+5lamda+1))/(1+4+1)=-(1)/(6)`
`therefore h= lammda-(1)/(6) ,k =- 3lamda -(1)/(3) ,il=-5 lamda+(1)/(6)`
so, foot ro rpoint on lovus M is `(lamda-(1)/(6),-3lamda-(1)/(3)-(1)/(3),-5lamda+(1)/(6))`
So points (a) and (b) lie on this loucs .