The shortest distance between the lines x + 1 = 2y = - 12z and x =...

Given,

Equation of the lines

$x + 1 = 2y = - 12z$ and $x = y + 2 = 6z - 6$

$⇒ \dfrac{x + 1}{1} = \dfrac{y}{\dfrac{1}{2}} = \dfrac{z}{\dfrac{- 1}{12}}$ and $\dfrac{x}{1} = \dfrac{y + 2}{1} = \dfrac{z - 1}{\dfrac{1}{6}}$

We know that, shortest distance between the lines is given by $S . D = \dfrac{\left(\overset{\to}{b} - \overset{\to}{a}\right) \cdot \left(\overset{\to}{p} \times \overset{\to}{q}\right)}{\left|\overset{\to}{p} \times \overset{\to}{q}\right|}$

Here \overset{\to}{p} = \overset{^}{i} + \overset{^}{j}2 + \dfrac{- \overset{^}{k}}{12} \text{ and } \overset{\to}{q} = \overset{^}{i} + \overset{^}{j} + \overset{^}{k}6 and \overset{\to}{a} = \overset{^}{i} \text{ and } \overset{\to}{b} = 2\overset{^}{j} - \overset{^}{k}

⇒ S . D = \left(- \overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right) \cdot \dfrac{\left(\overset{\to}{p} \times \overset{\to}{q}\right)}{\left|\overset{\to}{p} \times \overset{\to}{q}\right|}

Now solving \overset{\to}{p} \times \overset{\to}{q} ≡ \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & \dfrac{1}{2} & \dfrac{- 1}{12} \\ 1 & 1 & \dfrac{1}{6}\end{cases}\right|

⇒ \overset{\to}{p} \times \overset{\to}{q} ≡ \dfrac{1}{6}\overset{^}{i} - \dfrac{1}{4}\overset{^}{j} + \dfrac{1}{2}\overset{^}{k}

⇒ \overset{\to}{p} \times \overset{\to}{q} ≡ 2\overset{^}{i} - 3\overset{^}{j} + 6\overset{^}{k}

Hence, S . D = \dfrac{\left(- \overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right) \cdot \left(2\overset{^}{i} - 3\overset{^}{j} + 6\overset{^}{k}\right)}{\sqrt{{2}^{2} + {3}^{2} + {6}^{2}}} = \left|\dfrac{- 14}{7}\right| = 2