NCERT Class 12 · Ch 11 Exercise 11.2 · Q14
NCERT Class 12 Maths · Chapter 11 Three Dimensional Geometry · Exercise 11.2 Question 14
Find the shortest distance between r=(1,2,3)+t(1,−3,2) and r=(4,5,6)+s(2,3,1).
💡 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: 3*sqrt(19)/19
- For nonparallel directions, D=|(B−A)·(d×e)|/|d×e|.
- Here B−A=(3,3,3), d×e=(-9,3,9).
- The numerator is 9 and the denominator is 3*sqrt(19).
- Therefore D=3*sqrt(19)/19.
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