TS EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the skew lines \(\frac{x-2}{1}=\frac{y-3}{-2}=\frac{x+5}{1}\) and \(\frac{x-1}{-1}=\frac{y+2}{3}=\frac{z-4}{2}\) is
- A \(\frac{22}{\sqrt{59}}\)
- B \(\frac{21}{\sqrt{59}}\)
- C \(\frac{31}{\sqrt{59}}\)
- D \(31 \sqrt{59}\)
Answer & Solution
Correct Answer
(C) \(\frac{31}{\sqrt{59}}\)
Step-by-step Solution
Detailed explanation
Given lines are \(\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z+5}{1}\) and \(\frac{x-1}{-1}=\frac{y+2}{3}=\frac{z-4}{2}\) Here, \(\begin{aligned} & x_1=2, y_1=3, z_1=-5 \\ & a_1=1, b_1=-2, c_1=1 \\ & x_2=1, y_2=-2, z_2=4 \\ & a_2=-1, b_2=3, c_2=2\end{aligned}\) \(\therefore\) Shortest…
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