The shortest distance between the lines \(\vec{r} = (\hat{\imath}-\hat{\jmath})+\lambda(\hat{\imath}+2\hat{\jmath}+2\hat{k})\) and \(\vec{r} = (2\hat{\imath}+\hat{\jmath}+3\hat{k})+\mu(2\hat{\imath}-2\hat{\jmath}+\hat{k})\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{2}{3}\)
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
\( \displaystyle d = \frac{|(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)|}{|\vec{b}_1\times\vec{b}_2|} \)
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 2 \\ 2 & -2 & 1 \end{vmatrix}\)
\(= \hat{\imath}(2+4)-\hat{\jmath}(1-4)+\hat{k}(-2-4) = 6\hat{\imath}+3\hat{\jmath}-6\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{36+9+36} = 9\)
\(\vec{a}_2-\vec{a}_1 = \hat{\imath}+2\hat{\jmath}+3\hat{k}\)
\((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = 6+6-18 = -6\)
\( \displaystyle d = \frac{|-6|}{9} = \frac{2}{3} \)
Why the other options are incorrect
Option (2) 6 — Incorrect. This is only the numerator \(|-6|\), not divided by \(|\vec{b}_1\times\vec{b}_2| = 9\).
Option (3) \(\displaystyle\frac{2}{27}\) — Incorrect. This divides 6 by \(|\vec{b}_1\times\vec{b}_2|^2 = 81\) instead of 9.
Option (4) 2 — Incorrect. This divides 6 by \(|\vec{b}_1| = 3\) instead of \(|\vec{b}_1\times\vec{b}_2| = 9\).