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Question

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+2yz+1=0 and P2:2xy+z1=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?

A
(56,0,16)
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B
(16,13,16)
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C
(0,56,23)
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D
(13,0,23)
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Solution

The correct option is C (0,56,23)
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

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+133=z165

where ^i3^j5^k is vector parallel to line of intersection of P1 and P2.

On checking (0,56,23), (16,13,16) satisfy the above equation.

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