NCERT Class 12 · Ch 11 Exercise 11.2 · Q12
NCERT Class 12 Maths · Chapter 11 Three Dimensional Geometry · Exercise 11.2 Question 12
Find the shortest distance between r=(1,2,1)+t(1,−1,1) and r=(2,−1,−1)+s(2,1,2).
💡 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: 3*sqrt(2)/2
- For nonparallel directions, D=|(B−A)·(d×e)|/|d×e|.
- Here B−A=(1,-3,-2), d×e=(-3,0,3).
- The numerator is 9 and the denominator is 3*sqrt(2).
- Therefore D=3*sqrt(2)/2.
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