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← JEE Advanced 2013 · Paper 1
MathematicsMultiple correct+4 / -1Question 55 of 60

JEE Advanced 2013 · Paper 1, Mathematics question 55

A line ll passing through the origin is perpendicular to the lines
l1:(3+t)i^+(−1+2t)j^+(4+2t)k^,−∞<t<∞l_1:(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k,\quad-\infty<t<\infty
l2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<s<∞l_2:(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k,\quad-\infty<s<\infty
Then, the coordinate(s) of the point(s) on l2l_2 at a distance of 17\sqrt{17} from the point of intersection of ll and l1l_1 is (are)

Choose one or more options
Answer and worked solution

Official answer: B, D

Direction of l. Perpendicular to (1,2,2)(1,2,2) and (2,2,1)(2,2,1): (1,2,2)×(2,2,1)=(−2,3,−2)(1,2,2)\times(2,2,1)=(-2,3,-2).

Meeting l1l_1. (3+t,−1+2t,4+2t)=λ(−2,3,−2)(3+t,-1+2t,4+2t)=\lambda(-2,3,-2) gives t=−1t=-1, λ=−1\lambda=-1: the point (2,−3,2)(2,-3,2).

Points on l2l_2. (3+2s,3+2s,2+s)(3+2s,3+2s,2+s) at distance 17\sqrt{17}:
(1+2s)2+(6+2s)2+s2=17  ⟹  9s2+28s+20=0  ⟹  s=−2, −109.(1+2s)^2+(6+2s)^2+s^2=17\implies9s^2+28s+20=0\implies s=-2,\ -\tfrac{10}9.
These give (−1,−1,0)(-1,-1,0) and (79,79,89)\left(\frac79,\frac79,\frac89\right).

(B), (D)

Topics:Equation of a line in spaceCross productAngles and distances
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