Let the equation of lines be as \(L_1 = \vec{a_1}+\lambda\vec{b_1}\) and \(L_2 = \vec{a_2}+\lambda\vec{b_2}\) such that \(\vec{a_1}-\vec{a_2} = 2\hat{\imath}+4\hat{\jmath}+ 4\hat{k}\) and \(\vec{b_1}\times\vec{b_2} = 8\hat{\imath}-4\hat{k}\). Then the shortest distance between \(L_1\) and \(L_2\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
The lines pass through the points with position vectors \(\vec{a_1}\) and \(\vec{a_2}\) and are parallel to \(\vec{b_1}\) and \(\vec{b_2}\) (the parameter on each line varies independently). Since \(\vec{b_1}\times\vec{b_2} \ne \vec{0}\), the lines are not parallel, and the shortest distance between two lines is
\(\displaystyle d = \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|}{|\vec{b_1}\times\vec{b_2}|}\)
Here \(\vec{a_2}-\vec{a_1} = -(2\hat{\imath}+4\hat{\jmath}+4\hat{k})\), and the sign does not matter inside the modulus:
\((2\hat{\imath}+4\hat{\jmath}+4\hat{k})\cdot(8\hat{\imath}+0\hat{\jmath}-4\hat{k}) = 16 + 0 - 16\)
\(= 0\)
So d = 0: the lines intersect.