If the shortest distance between the line joining the points (1, 2,...

Equation of line joining $\left(1, 2, 3\right)$ and $\left(2, 3, 4\right)$ is$</p><p>$\overset{\to}{r} = \left(\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}\right) + \lambda \left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)

Also, vector equation is$</p><p>$\overset{\to}{r} = \left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) + \mu\left(2\overset{^}{i} - \overset{^}{j}\right)

So,$</p><p>$\overset{\to}{a}1 = \left(\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}\right)

\overset{\to}{a}2 = \left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right)

Hence,$</p><p>$\overset{\to}{a}2 - \overset{\to}{a}1 = 0\overset{^}{i} - 3\overset{^}{j} - \overset{^}{k}$</p><p>And,$

\overset{\to}{b}1 = \overset{^}{i} + \overset{^}{j} + \overset{^}{k}

\overset{\to}{b}2 = 2\overset{^}{i} - \overset{^}{j} + 0\overset{^}{k}

And, $</p><p>$\overset{\to}{b}1 \times \overset{\to}{b}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0\end{cases}\right|= \overset{^}{i} + 2\overset{^}{j} - 3\overset{^}{k}

$⇒ \left|\overset{\to}{b}1 \times \overset{\to}{b}2\right|$ = \sqrt{14}$</p><p>Required shortest distance is$

$\alpha = \dfrac{\left|\left(\overset{\to}{a}2 - \overset{\to}{a}1\right) \cdot \left(\overset{\to}{b}1 \times \overset{\to}{b}2\right)\right|}{\left|\overset{\to}{b}1 \times \overset{\to}{b}2\right|}$

⇒ \alpha = \left|\dfrac{\left(- 3\overset{^}{j} - \overset{^}{k}\right) \cdot \left(\overset{^}{i} + 2\overset{^}{j} - 3\overset{^}{k}\right)}{\sqrt{14}}\right|

$⇒ \alpha = \left|\dfrac{- 6 + 3}{\sqrt{14}}\right| = \dfrac{3}{\sqrt{14}}$

Now, $28{\alpha}^{2} = 228 \times \dfrac{9}{14} = 18$.