The shortest distance between the lines x = y + 2 = 6 z - 6 and x + 1 = 2 y = - 12 z is:

Question

The shortest distance between the lines $x = y + 2 = 6 z - 6$ and $x + 1 = 2 y = - 12 z$ is:

Smart Achievers · Accepted Answer

(2)

Ans. 2

Explanation:

The lines are $\frac{x}{6} = \frac{y + 2}{6} = \frac{z - 1}{1}$

and $\frac{x + 1}{12} = \frac{y}{6} = \frac{z}{- 1}$

Here, ${\vec{a}}_{1} = - 2 \hat{j} + \hat{k}, {\vec{b}}_{1} = 6 \hat{i} + 6 \hat{j} + \hat{k},$

${\vec{a}}_{2} = - \hat{i},$

${\vec{b}}_{2} = 12 \hat{i} + 6 \hat{j} - \hat{k}$

${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 6 & 1 \\ 12 & 6 & - 1 \end{matrix} |$

$= - 12 \hat{i} + 18 \hat{j} - 36 \hat{k}$

$Shortest distance =$ $\frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$

$= \frac{| (- \hat{i} + 2 \hat{j} - \hat{k}) \cdot (- 12 \hat{i} + 18 \hat{j} - 36 \hat{k}) |}{\sqrt{{(- 12)}^{2} + {(18)}^{2} + {(- 36)}^{2}}}$

$= \frac{| 12 + 36 + 36 |}{\sqrt{1764}} = \frac{84}{42} = 2$

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