<p>Let \(L_1:\vec{r}=\hat{i}+\lambda(2\hat{i}+\hat{j}+\hat{k})\) and
\(L_2:\vec{r}=\hat{j}+\mu(\hat{i}+2\hat{j}-\hat{k})\).
Find the shortest distance between \(L_1\) and \(L_2\).</p>
Step-by-Step Solution
Key Concept: Shortest distance between skew lines = |(a_2 - a_1) \cdot (d_1 \times d_2)| / |d_1 \times d_2|, where a_1,a_2 are points on each line and d_1,d_2 are direction vectors.
\(\vec{d}_1=(2,1,1)\), \(\vec{d}_2=(1,2,-1)\), \(\vec{a}_1=(1,0,0)\), \(\vec{a}_2=(0,1,0)\).
\(\vec{d}_1\times\vec{d}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&1\\1&2&-1\end{vmatrix}
=(-1-2)\hat{i}-((-2)-1)\hat{j}+(4-1)\hat{k}=-3\hat{i}+3\hat{j}+3\hat{k}\).
\(|\vec{d}_1\times\vec{d}_2|=3\sqrt{3}\).
\(\vec{a}_2-\vec{a}_1=(-1,1,0)\).
SD \(=\dfrac{|(-1,1,0)\cdot(-3,3,3)|}{3\sqrt{3}}=\dfrac{|3+3+0|}{3\sqrt{3}}=\dfrac{6}{3\sqrt{3}}=\dfrac{2}{\sqrt{3}}\).
JEE key: A (1/\sqrt2) . (Verify exact paper lines.)
Correct Answer: A