NCERT Class 12 · Ch 11 Exercise 11.2 · Q15
NCERT Class 12 Maths · Chapter 11 Three Dimensional Geometry · Exercise 11.2 Question 15
Find the shortest distance between r=(1−t,t−2,3−2t) and r=(s+1,2s−1,−2s−1).
💡 Concept Hints
- Identify the base point and a nonzero direction before applying a vector formula.
- Use dot products for angles, proportional directions for parallelism, and a triple product for skew-line distance.
✅ Step-by-Step Solution
Final Answer: 8*sqrt(29)/29
- For nonparallel directions, D=|(B−A)·(d×e)|/|d×e|.
- Here B−A=(0,1,-4), d×e=(2,-4,-3).
- The numerator is 8 and the denominator is sqrt(29).
- Therefore D=8*sqrt(29)/29.
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