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

The line `L` passes through the origin and is parallel to theplanes `P_(1)` and `P_(2)`.
Let the equation of line `L` be `(x-0)/a=(y-0)/b=(z-0)/c`
It is parallel to the planes `P_(1)` and `P_(2)`.
`:.a+2b-c=0` an `2a-b+c=0`
`impliesa/1=b/(-3)=c/(-5)impliesa/(-1)=b/3=c/5`
So the equation of line `L` is
`x/(-1)=y/3=z/5`.............i
The coordinates of ay pints on line(i) are `(-gamma,3gamma,5gamma)`. The coordinates of the foot of perpendicular from `(-gamma,3gamma,5gamma)` on the plane `P_(1)` are given by
`(x+gamma)/1=(y-3gamma)/2=(3-5gamma)/(-1)=(-gamma+6gamma-1)/(1+4+1)`
or `(x+gamma)/1=(y-3gamma)/2=(3-5gamma)/(-1)=-1/6`
`impliesx=-gamma=1/5,y=3gamma,z=5gamma+1/6`
So, the coordinats of the foot of perpendicular from `(-gamma,3gamma,5gamma)` are `(-gamma=1/6,3gamma-2/6,5gamma+1/6)`.
The clearly `M` is the locus of points `(-gamma-1/6,3gamma-2/6,5gamma+1/6)`, where `gamma` is a variable. We observe that the coordinates on options (a) and (b) satisfy for `gamma=1/6` and `gamma=0` respectively.