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← JEE Advanced 2015 · Paper 1
MathematicsMultiple correct+4 / -2Question 52 of 60

JEE Advanced 2015 · Paper 1, Mathematics question 52

In R3\mathbb R^3, let LL be a straight line passing through the origin. Suppose that all the points on LL are at a constant distance from the two planes P1:x+2y−z+1=0P_1:x+2y-z+1=0 and P2:2x−y+z−1=0P_2:2x-y+z-1=0. Let MM be the locus of the feet of the perpendiculars drawn from the points on LL to the plane P1P_1. Which of the following points lie(s) on MM?

Choose one or more options
Answer and worked solution

Official answer: A, B

Direction of L. Points of L are equidistant from both planes, so L is parallel to both:
(1,2,−1)×(2,−1,1)=(1,−3,−5).(1,2,-1)\times(2,-1,1)=(1,-3,-5).
So L is t(1,−3,−5)t(1,-3,-5).

Feet on P1P_1. For a point on L, x+2y−z+1=t(1−6+5)+1=1x+2y-z+1=t(1-6+5)+1=1. The foot is
(t−16, −3t−13, −5t+16).\left(t-\tfrac16,\ -3t-\tfrac13,\ -5t+\tfrac16\right).

Check the points.

  • (A) t=16t=\frac16 gives (0,−56,−23)\left(0,-\frac56,-\frac23\right). ✓
  • (B) t=0t=0 gives (−16,−13,16)\left(-\frac16,-\frac13,\frac16\right). ✓
  • (C), (D) No tt fits.

(A), (B)

Topics:Angles and distancesEquation of a line in space
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