TS EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the skew-lines \(\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}, \frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}\) is
- A 6
- B 7
- C \(3 \sqrt{3}\)
- D \(\sqrt{35}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{35}\)
Step-by-step Solution
Detailed explanation
Equations of given lines in vector form can be written as…
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