In R 3 , let L be a straight line passing through the origin. Suppose that all the points on L are at a…

In R3 , 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 P1 :x+2y-z+1=0  and P2 :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 P1 . Which of the following points lie (s) on M ?
  1. 0, -56, -23
  2. -16, -13,16
  3. -56, 0,16 
  4. -13, 0,23

Solution

Line L will lie on one of the angle bisector planes and will be parallel to line of intersection of the given planes. Also locus of foot of perpendicular of line L in plane P1 will be line M, which will be parallel L.
Foot of perpendicular from (0, 0, 0) on plane P1 is -16, -13,16
Hence equation of line M
x+161=y+13-3=z-16-5
Where i^-3j^-5k^ is vector parallel to line of intersection of P1 and P2 .
On checking 0, -56-23, -16, -13,16 satisfy the above equation.

Asked in: JEE Advanced 2015 (Paper 1)

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