MHT CET · Maths · Three Dimensional Geometry
The shortest distance between the lines \(\frac{x+1}{3}=\frac{y-2}{2}=\frac{z+1}{2}\) and \(\frac{x-2}{1}=\frac{y-2}{2}=\frac{z+3}{3}\) is
- A \(\frac{2}{\sqrt{69}}\) units
- B \(\frac{14}{\sqrt{69}}\) units
- C \(\frac{9}{\sqrt{69}}\) units
- D \(\frac{1}{\sqrt{69}}\) units
Answer & Solution
Correct Answer
(A) \(\frac{2}{\sqrt{69}}\) units
Step-by-step Solution
Detailed explanation
\( \vec{a_1} = (-1, 2, -1), \vec{b_1} = (3, 2, 2) \) \( \vec{a_2} = (2, 2, -3), \vec{b_2} = (1, 2, 3) \)
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