Let the equation of lines be as $L_1 = \vec{a_1}+λ\vec{b_1}$ and $L_2 = \vec{a_2}+λ\vec{b_2}$ such that $\vec{a_1}-\vec{a_2} = 2\hat i+4\hat j+ 4\hat k$ and $\vec{b_1}×\vec{b_2} = 8\hat i-4\hat k$. Then the shortest distance between $L_1$ and $L_2$ is

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The correct answer is Option (3) → 0

Lines: $\vec{r}=\vec{a}_1+\lambda\vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu\vec{b}_2$.

Given: $\vec{a}_1-\vec{a}_2=\langle 2,4,4\rangle$, $\ \vec{b}_1\times\vec{b}_2=\langle 8,0,-4\rangle$.

Shortest distance formula: $\displaystyle d=\frac{|(\vec{a}_1-\vec{a}_2)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|}$.

Numerator: $(2,4,4)\cdot(8,0,-4)=16+0-16=0$.

$\Rightarrow d=\frac{|0|}{|\langle 8,0,-4\rangle|}=0$.

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