The shortest distance between the lines $\vec r=(4\hat i-\hat j)+\lambda(\hat i+2\hat j-3\hat k)$ and $\vec r=(\hat i-\hat j+2\hat k)+\mu(2\hat i+4\hat j-5\hat k)$ is

Direction vectors: $ \vec b_1=(1,2,-3),\ \vec b_2=(2,4,-5) $ $ \vec b_1\times \vec b_2=(2,-1,0) $ $ |\vec b_1\times \vec b_2|=\sqrt{4+1}=\sqrt{5} $ Points: $ \vec a_1=(4,-1,0),\ \vec a_2=(1,-1,2) $ $ \vec a_2-\vec a_1=(-3,0,2) $ Shortest distance: $ d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times \vec b_2)|}{|\vec b_1\times \vec b_2|} $ $ =\frac{|(-3,0,2)\cdot(2,-1,0)|}{\sqrt{5}} $ $ =\frac{|-6|}{\sqrt{5}}=\frac{6}{\sqrt{5}} $