The shortest distance between the lines is: $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and…
The shortest distance between the lines is:
$\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$
$\frac{8\sqrt{29}}{29}$ units
$ $\frac{8}{29}$ \text{ units} $
$\frac{\sqrt{7}}{25}$ units
None of these
Solution
Given equations are $\vec{r} = (i - 2j - k) + t(-i + j - 2k)$ and $\vec{r} = (i - j - k) + s(i + 2j - 2k)$.
On comparing with $\vec{r} = \vec{a} + \lambda \vec{b}$, we have
$\begin{aligned}
\vec{a_1} = i - 2j + 3k, &\quad \vec{b_1} = -i + j - 2k, \\
\vec{a_2} = i - j - k, &\quad \vec{b_2} = i + 2j - 2k.
\end{aligned}$
The shortest distance between given lines is
$
\text{Shortest distance} = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right|.
$
Therefore, $\vec{a_2} - \vec{a_1} = i - j - k - (i - 2j + 3k) = 0i + j - 4k$.
$
\vec{b_1} \times \vec{b_2} = \begin{vmatrix} i & j & k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = 2i - 4j - 3k.
$
Therefore, $|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{29}$.
Therefore, the required shortest distance is
$
\text{Shortest distance} = \left| \frac{(2i - 4j - 3k) \cdot (0i + j - 4k)}{\sqrt{29}} \right| = \left| \frac{0 - 4 + 12}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}} = \frac{8\sqrt{29}}{29} \text{ units}.$