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Class 12 Three Dimensional Geometry · Long Answer3 Marksshortest_distance

Class 12 Maths · Three Dimensional Geometry · Long Answer #49

Find the shortest distance between r=(1,1,0)+t(2,-1,1), t∈ℝ and r=(2,1,-1)+t(3,-5,2), t∈ℝ.

✅ Step-by-Step Solution

Final Answer: 10*sqrt(59)/59
  1. For nonparallel lines use D=|(B−A)·(d×e)|/|d×e|.
  2. B−A=(1,0,-1), d×e=(3,-1,-7).
  3. The numerator is 10 and the denominator is sqrt(59).
  4. Therefore D=10*sqrt(59)/59.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] For nonparallel lines use D=|(B−A)·(d×e)|/|d×e|.
  • [1 mark] B−A=(1,0,-1), d×e=(3,-1,-7).
  • [1 mark] The numerator is 10 and the denominator is sqrt(59).
  • [0.5 marks] Therefore D=10*sqrt(59)/59.

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