The distance between the parallel lines \(\vec{r} = (\hat{\imath}+2\hat{\jmath}-\hat{k})+\lambda(2\hat{\imath}-\hat{\jmath}+2\hat{k})\) and \(\vec{r} = (3\hat{\imath}-\hat{\jmath}+\hat{k})+\mu(2\hat{\imath}-\hat{\jmath}+2\hat{k})\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{4\sqrt{2}}{3}\)
For the parallel lines \(\vec{r} = \vec{a}_1+\lambda\vec{b}\) and \(\vec{r} = \vec{a}_2+\mu\vec{b}\), the distance is
\( \displaystyle d = \frac{|\vec{b}\times(\vec{a}_2-\vec{a}_1)|}{|\vec{b}|} \)
Here \(\vec{b} = 2\hat{\imath}-\hat{\jmath}+2\hat{k}\) and
\(\vec{a}_2-\vec{a}_1 = 2\hat{\imath}-3\hat{\jmath}+2\hat{k}\)
\(\vec{b}\times(\vec{a}_2-\vec{a}_1) = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 2 \\ 2 & -3 & 2 \end{vmatrix}\)
\(= \hat{\imath}(-2+6)-\hat{\jmath}(4-4)+\hat{k}(-6+2) = 4\hat{\imath}-4\hat{k}\)
\(|4\hat{\imath}-4\hat{k}| = \sqrt{32} = 4\sqrt{2}\)
\(|\vec{b}| = \sqrt{4+1+4} = 3\)
\( \displaystyle d = \frac{4\sqrt{2}}{3} \)
Why the other options are incorrect
Option (1) \(4\sqrt{2}\) — Incorrect. This is \(|\vec{b}\times(\vec{a}_2-\vec{a}_1)|\) before dividing by \(|\vec{b}| = 3\).
Option (2) \(\displaystyle\frac{11}{3}\) — Incorrect. This uses the dot product \(\vec{b}\cdot(\vec{a}_2-\vec{a}_1) = 4+3+4 = 11\) in place of the cross product.
Option (4) \(\displaystyle\frac{4\sqrt{2}}{9}\) — Incorrect. This divides by \(|\vec{b}|^2 = 9\) instead of \(|\vec{b}| = 3\).