The shortest distance between the lines \(\vec{r} = (\hat{\imath} + 2\hat{\jmath} + 3\hat{k}) + \lambda(2\hat{\imath} + 3\hat{\jmath}+4\hat{k})\) and \(\vec{r} = (2\hat{\imath} + 4\hat{\jmath} + 5\hat{k}) + \mu(3\hat{\imath} + 4\hat{\jmath} + 5\hat{k})\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{\sqrt{6}}\)
For the lines \(\vec{r} = \vec{a}_1 + \lambda\vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}_2\), the shortest distance is
\( \displaystyle d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right| \)
Here
\(\vec{b}_1 = 2\hat{\imath} + 3\hat{\jmath} + 4\hat{k},\quad \vec{b}_2 = 3\hat{\imath} + 4\hat{\jmath} + 5\hat{k}\)
\(\vec{a}_2 - \vec{a}_1 = \hat{\imath} + 2\hat{\jmath} + 2\hat{k}\)
\( \displaystyle \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&3&4\\3&4&5\end{vmatrix} \)
\(= \hat{\imath}(15 - 16) - \hat{\jmath}(10 - 12) + \hat{k}(8 - 9)\)
\(= -\hat{\imath} + 2\hat{\jmath} - \hat{k}\)
\(|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 4 + 1} = \sqrt{6}\)
\((\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-1)(1) + (2)(2) + (-1)(2) = 1\)
So
\( \displaystyle d = \frac{1}{\sqrt{6}} \)