The shortest distance between the lines \(\frac{x−3}{2}=\frac{y−2}{3}=\frac{z−1}{−1}\) and \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z−5}{3}\) is
- \(\frac{18}{\sqrt5}\)
- \(\frac{22}{3\sqrt5}\)
- \(\frac{46}{3\sqrt5}\)
\(6\sqrt3\)
The Correct Option is A
Solution and Explanation
\(L_1\) : \(\frac{x−3}{2}=\frac{y−2}{3}=\frac{z−1}{−1}\)
\(L_2\) : \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z−5}{3}\)
Now,
\(\overrightarrow p×\overrightarrow q\) = \(\begin{vmatrix}\hat i& \hat j& \hat k\\ 2 &3 &−1 \\ 2 & 1& 3 \end{vmatrix} =10\hat i−8\hat j−4\hat k\)
\(\overrightarrow a_2−\overrightarrow a_1=6\hat i−4\hat j−4\hat k\)
\(∴S.D=\begin{vmatrix}\frac{60+32+16}{\sqrt{100+64+16}}\end{vmatrix}\)
= \(\frac{108}{\sqrt{180}}=\frac{18}{\sqrt5}\)
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Concepts Used:
Shortest Distance Between Two Parallel Lines
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)