Question

Find the distance of the line \( \frac{x-1}{2}=\frac{y-1}{-1}=\frac{z}{1} \) from the line parallel to it and passing through \( (2,1,-1) \).

Answer 1

Distance between two lines is

d = \(|\frac{\vec b\times(\vec a_2-\vec a_1)}{|\vec b|}|\) 

Here \(\vec {a_1}=\hat i+\hat j + 0\hat k\) (\(\because\) line (1) passing through point (1, 1, 0))

\(\vec {a_2}=2\hat i+\hat j-\hat k\) (\(\because\) line (2) is passing through point (2, 1, -1))

And \(\vec b=2\hat i-\hat j+\hat k\) (\(\vec b\) is direction vector of both lines whose direction cosines are 2, -1, 1)

\(\therefore\) \(\vec a_2 - \vec a_1=(2\hat i+\hat j-\hat k)-(\hat i+\hat j+0\hat k)\) 

 = \(\hat i+0\hat j-\hat k\) 

\(\therefore\) \(\vec b\times(\vec a_2-\vec a_1)=(2\hat i-\hat j+\hat k).(\hat i+0\hat j-\hat k)\) 

 = 2 x 1 - 1 x 0 + 1 x -1

 = 2 - 1 = 1

and  |\(\vec b\)| = |\(2\hat i-\hat j+\hat k\)| = \(\sqrt{2^2+(-1)^2+1^2}\) = \(\sqrt{4+1+1}\) = \(\sqrt6\) 

\(\therefore\) d₁ = \(|\frac{\vec b\times(\vec a_2-\vec a_1)}{|\vec b|}|\) = \(\frac{1}{\sqrt6}\) = \(\frac{\sqrt6}6\) units