Three Dimensional Geometry
RS Aggarwal
CBSE
Grade 12
Question:
Find the shortest distance between lines $\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}$.
Step-by-Step Solution
Key Concept: d = 2\sqrt{29}.
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Correct Answer:
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