The shortest distance between the following lines: $\vec r=(\hat i+\hat j−\hat k)+s(2\hat i+\hat j+ \hat k), \vec r= (\hat i+\hat j+2\hat k) +t(4\hat i +2\hat j +2\hat k)$, where $s$ and $t$ are scalars, is:

$\sqrt{\frac{15}{2}}$

The correct answer is Option (2) → $\sqrt{\frac{15}{2}}$

Let the lines be:

Line 1: $\vec{r}_1 = (i + j - k) + s(2i + j + k)$

Line 2: $\vec{r}_2 = (i + j + 2k) + t(4i + 2j + 2k)$

Direction vector of Line 1: $\vec{a} = \langle 2, 1, 1 \rangle$

Direction vector of Line 2: $\vec{b} = \langle 4, 2, 2 \rangle$

Since $\vec{b} = 2\vec{a}$, the lines are parallel.

Let $\vec{A} = \langle 1, 1, -1 \rangle$ (point on Line 1)

Let $\vec{B} = \langle 1, 1, 2 \rangle$ (point on Line 2)

Then $\vec{AB} = \vec{B} - \vec{A} = \langle 0, 0, 3 \rangle$

Since lines are parallel, shortest distance is the perpendicular distance from point on one line to the other line.

Find unit direction vector $\vec{d}$ of line: $\vec{d} = \frac{\vec{a}}{|\vec{a}|} = \frac{\langle 2, 1, 1 \rangle}{\sqrt{6}}$

Now, shortest distance between parallel lines is:

$D = \frac{|\vec{AB} \times \vec{a}|}{|\vec{a}|}$

$\vec{AB} = \langle 0, 0, 3 \rangle$, $\vec{a} = \langle 2, 1, 1 \rangle$

$\vec{AB} \times \vec{a} = \begin{vmatrix} i & j & k \\ 0 & 0 & 3 \\ 2 & 1 & 1 \end{vmatrix} = i(0 - 3) - j(0 - 6) + k(0 - 0) = -3i + 6j$

$|\vec{AB} \times \vec{a}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}$

$|\vec{a}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$

So, shortest distance:

$D = \frac{\sqrt{45}}{\sqrt{6}} = \sqrt{\frac{45}{6}}= \sqrt{\frac{15}{2}}$