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NDA 2020 Maths · Mathematics · Three Dimensional Geometry · hard

The shortest distance between the lines vec r=(1,2,3)+t(1,1,0) and vec r=(3,3,6)+s(0,1,1) is

Single correct option

The shortest distance between the lines r⃗=(1,2,3)+t(1,1,0)\vec r=(1,2,3)+t(1,1,0) and r⃗=(3,3,6)+s(0,1,1)\vec r=(3,3,6)+s(0,1,1) is
  1. A. 43\dfrac43
  2. B. 23\dfrac{2}{\sqrt3}
  3. C. 43\dfrac{4}{\sqrt3}
  4. D. 44
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Answer: (C) 43\dfrac{4}{\sqrt3}

The common perpendicular direction is n⃗=(1,1,0)×(0,1,1)=(1,−1,1)\vec n=(1,1,0)\times(0,1,1)=(1,-1,1), with ∣n⃗∣=3|\vec n|=\sqrt3. The vector joining the given points is (2,1,3)(2,1,3), and its dot product with n⃗\vec n is 2−1+3=42-1+3=4. Shortest distance =∣4∣3=\dfrac{|4|}{\sqrt3}. The value 44 forgets to divide by ∣n⃗∣|\vec n|, and 43\dfrac43 divides by ∣n⃗∣2|\vec n|^2 instead.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake