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GUJCET · Maths · Three Dimensional Geometry
The shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\) and \(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}\) is ___________ .
- A \(\sqrt{\frac{209}{49}}\)
- B \(\sqrt{\frac{293}{49}}\)
- C \(\sqrt{\frac{209}{7}}\)
- D \(\sqrt{\frac{293}{7}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{293}{49}}\)
Step-by-step Solution
Detailed explanation
\( \vec{a_1} = (1, 2, -4), \vec{a_2} = (3, 3, -5), \vec{b} = (2, 3, 6) \) \( \vec{a_2} - \vec{a_1} = (3-1, 3-2, -5-(-4)) = (2, 1, -1) \)
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