If \( \vec{r}=\vec{a_{1}}+\lambda\vec{b_{1}} \) and \( \vec{r}=\vec{a_{2}}+\mu\vec{b_{2}} \) are two skew lines such that \( \vec{b_{1}}\times\vec{b_{2}}=2\hat{\imath}+3\hat{\jmath}+2\hat{k},\ \vec{a_{1}}=\hat{\imath}+\hat{\jmath}+\hat{k} \) and \( \vec{a_{2}}=2\hat{\imath}+3\hat{\jmath}+\hat{k} \), then the shortest distance between the lines is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{8}{\sqrt{17}} \)
The shortest distance between skew lines is
\( \displaystyle d=\frac{\left|(\vec{a_{2}}-\vec{a_{1}})\cdot(\vec{b_{1}}\times\vec{b_{2}})\right|}{\left|\vec{b_{1}}\times\vec{b_{2}}\right|} \)
Here
\( \vec{a_{2}}-\vec{a_{1}}=\hat{\imath}+2\hat{\jmath} \)
\( (\vec{a_{2}}-\vec{a_{1}})\cdot(\vec{b_{1}}\times\vec{b_{2}})=(1)(2)+(2)(3)+(0)(2)=8 \)
\( \left|\vec{b_{1}}\times\vec{b_{2}}\right|=\sqrt{4+9+4}=\sqrt{17} \)
So
\( \displaystyle d=\frac{8}{\sqrt{17}} \)