The shortest distance between the lines whose vector equations are \( \vec{r}=(\hat{\imath}+2\hat{\jmath}+3\hat{k})+\lambda(2\hat{\imath}+3\hat{\jmath}+4\hat{k}) \) and \( \vec{r}=(2\hat{\imath}+4\hat{\jmath}+5\hat{k})+\mu(4\hat{\imath}+6\hat{\jmath}+8\hat{k}) \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\sqrt{5}}{\sqrt{29}} \)
The direction \( 4\hat{\imath}+6\hat{\jmath}+8\hat{k}=2(2\hat{\imath}+3\hat{\jmath}+4\hat{k}) \), 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 \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+4\hat{k} \) and \( \vec{a}_{2}-\vec{a}_{1}=\hat{\imath}+2\hat{\jmath}+2\hat{k} \).
\( \displaystyle \vec{b}\times(\vec{a}_{2}-\vec{a}_{1})=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & 3 & 4\\ 1 & 2 & 2\end{vmatrix} \)
\( \displaystyle =(6-8)\hat{\imath}-(4-4)\hat{\jmath}+(4-3)\hat{k} \)
\( \displaystyle =-2\hat{\imath}+\hat{k} \)
\( \displaystyle |\vec{b}\times(\vec{a}_{2}-\vec{a}_{1})|=\sqrt{4+0+1}=\sqrt{5} \)
\( \displaystyle |\vec{b}|=\sqrt{4+9+16}=\sqrt{29} \)
\( \displaystyle d=\frac{\sqrt{5}}{\sqrt{29}} \)
Tip: Check whether the lines are parallel first: the skew-lines formula with \( \vec{b}_{1}\times\vec{b}_{2} \) fails here because that cross product is the zero vector.