NCERT Class 12 · Ch 11 Miscellaneous Exercise · Q4
NCERT Class 12 Maths · Chapter 11 Three Dimensional Geometry · Miscellaneous Exercise Question 4
Find the shortest distance between r=(6,2,2)+t(1,−2,2) and r=(−4,0,−1)+s(3,−2,−2).
💡 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: 9
- For nonparallel directions, D=|(B−A)·(d×e)|/|d×e|.
- Here B−A=(-10,-2,-3), d×e=(8,8,4).
- The numerator is 108 and the denominator is 12.
- Therefore D=9.
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