The shortest distance between the lines \(\vec r= \hat{\imath}+\hat{\jmath}+\lambda(2\hat{\imath}-\hat{\jmath}+\hat{k})\) and \(\vec r = 2\hat{\imath}+\hat{\jmath}-\hat{k}+\mu(4\hat{\imath}-2\hat{\jmath} + 2\hat{k})\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\sqrt{66}}{6}\)
The direction vectors are \(2\hat{\imath} - \hat{\jmath} + \hat{k}\) and \(4\hat{\imath} - 2\hat{\jmath} + 2\hat{k} = 2(2\hat{\imath} - \hat{\jmath} + \hat{k})\), so the lines are parallel. For parallel lines \(\vec r = \vec a_1 + \lambda\vec b\) and \(\vec r = \vec a_2 + \mu\vec b\),
\(\displaystyle d = \frac{|\vec b \times (\vec a_2 - \vec a_1)|}{|\vec b|}\)
Here \(\vec b = 2\hat{\imath} - \hat{\jmath} + \hat{k}\), \(\vec a_1 = \hat{\imath} + \hat{\jmath}\) and \(\vec a_2 = 2\hat{\imath} + \hat{\jmath} - \hat{k}\):
\(\vec a_2 - \vec a_1 = \hat{\imath} - \hat{k}\)
\(\displaystyle \vec b \times (\vec a_2 - \vec a_1) = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&-1&1\\1&0&-1\end{vmatrix}\)
\(= (1 - 0)\hat{\imath} - (-2 - 1)\hat{\jmath} + (0 + 1)\hat{k}\)
\(= \hat{\imath} + 3\hat{\jmath} + \hat{k}\)
\(|\vec b \times (\vec a_2 - \vec a_1)| = \sqrt{1 + 9 + 1} = \sqrt{11}\)
\(|\vec b| = \sqrt{4 + 1 + 1} = \sqrt{6}\)
So
\(\displaystyle d = \frac{\sqrt{11}}{\sqrt{6}} = \frac{\sqrt{66}}{6}\)
Tip: Check first whether the direction vectors are proportional. For parallel lines the skew-lines formula \(\frac{|(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|}\) fails, because \(\vec b_1 \times \vec b_2 = \vec 0\).