The shortest distance between the lines $\vec r = (\hat i + 2\hat j + 3\hat k) + λ(2\hat i + 3\hat j+4\hat k)$ and $\vec r = (2\hat i + 4\hat j + 5\hat k) + μ(3\hat i + 4\hat j + 5\hat k)$ is equal

$\frac{1}{\sqrt{6}}$

The correct answer is Option (4) → $\frac{1}{\sqrt{6}}$

Given lines:

$\vec r_1 = \hat i+2\hat j+3\hat k + \lambda(2\hat i+3\hat j+4\hat k)$

$\vec r_2 = 2\hat i+4\hat j+5\hat k + \mu(3\hat i+4\hat j+5\hat k)$

Direction vectors:

$\vec a = \langle2,3,4\rangle,\;\vec b=\langle3,4,5\rangle$

Vector joining points:

$\vec c=\langle2-1,4-2,5-3\rangle=\langle1,2,2\rangle$

Cross product:

$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & 4\\ 3 & 4 & 5 \end{vmatrix}$

$=\hat i(15-16)-\hat j(10-12)+\hat k(8-9)$

$=\langle -1,2,-1\rangle$

Magnitude:

$|\vec a\times\vec b|=\sqrt{(-1)^{2}+2^{2}+(-1)^{2}}=\sqrt{6}$

Now

Shortest distance

$=\frac{|\vec c\cdot(\vec a\times\vec b)|}{|\vec a\times\vec b|}$

$\vec c\cdot(\vec a\times\vec b)=\langle1,2,2\rangle\cdot\langle-1,2,-1\rangle$

$=-1+4-2=1$

Hence distance

$=\frac{1}{\sqrt6}$