\(\vec r=2\hat i - 5\hat j+\hat k+\lambda(3\hat i+2\hat j+6\hat k)\)
\(\vec r=7\hat i - 6\hat k+\mu(\hat i+2\hat j+2\hat k)\)
\(\vec a_1=2\hat i-5\hat j+\hat k\), \(\vec b_1=3\hat i+2\hat j+6\hat k\)
\(\vec a_2=7\hat i-6\hat k\), \(\vec b_2=\hat i+2\hat j+2\hat k\)
distance between both lines is l = \(\left|\frac{(\vec a_2-\vec a_1).(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|\)
Now, \(\vec a_2-\vec a_1=(7\hat i-6\hat k)-(2\hat i-5\hat j+\hat k)\)
= \(5\hat i+5\hat j-7\hat k\)
\(\vec b_1\times\vec b_2\)\(=\begin{vmatrix}\hat i&\hat j&\hat k\\3&2&6\\1&2&3\end{vmatrix}\) = \(\hat i\begin{vmatrix}2&6\\2&2\end{vmatrix}\)\(-\hat j\begin{vmatrix}3&6\\1&2\end{vmatrix}\)\(+\hat k\begin{vmatrix}3&2\\1&2\end{vmatrix}\) = -8\(\hat i\) + 4\(\hat k\)
\((\vec a_2-\vec a_1).(\vec b_1\times\vec b_2)\) = (5\(\hat i\) + 5\(\hat j\) - 7\(\hat k\)). (-8\(\hat i\) + 0\(\hat j\) + 4\(\hat k\))
= 5 x -8 + 5 x 0 - 7 x 4
= -40 - 28 = -68
|\(\vec b_1\times \vec b_2\)| = |-8\(\hat i\) + 4\(\hat k\)| = \(\sqrt{8^2+4^2}\) = \(\sqrt{64+16}=\sqrt{80}\)
= \(\sqrt{16\times5}\) = \(4\sqrt5\)
\(\therefore\) d = \(\left|\frac{(\vec a_2-\vec a_1).(\vec b_1\times\vec b_2)}{|\vec b_1\times \vec b_2|}\right|\)
= \(|\frac{-68}{4\sqrt5}|\) = \(|\frac{-17}{\sqrt5}|\) = \(\frac{17}{\sqrt5}\) units
