First slide

Straight lines in 3D

Question

The distance between the parallel lines $\bar{r} = 2 \bar{i} + 3 \bar{j} - \bar{k} + λ (\bar{i} - \bar{j} + 2 \bar{k})$ and $\bar{r} = - 3 \bar{i} + 4 \bar{j} + \bar{k} + μ (\bar{i} - \bar{j} + 2 \bar{k})$ is

Moderate

A

$2 \sqrt{\frac{22}{3}}$
B

$\sqrt{\frac{175}{6}}$
C

$\frac{14}{\sqrt{6}}$
D

7

Solution

${\bar{a}}_{1} = 2 \bar{i} + 3 \bar{j} - \bar{k}$ , ${\bar{a}}_{2} = - 3 \bar{i} + 4 \bar{j} + \bar{k}$ , $\bar{b} = \bar{i} - \bar{j} + 2 \bar{k}$ , ${\bar{a}}_{1} - {\bar{a}}_{2} = 5 \bar{i} - \bar{j} - 2 \bar{k}$
$\bar{b} \times ({\bar{a}}_{1} - {\bar{a}}_{2}) = |\begin{matrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & - 1 & 2 \\ 5 & - 1 & - 2 \end{matrix}| = 4 \bar{i} + 12 \bar{j} + 4 \bar{k}$
$| \bar{b} \times ({\bar{a}}_{1} - {\bar{a}}_{2}) | = \sqrt{176}$ , $| \bar{b} | = \sqrt{6}$
Distance = $\frac{| \bar{b} \times ({\bar{a}}_{1} - {\bar{a}}_{2}) |}{| \bar{b} |} = \frac{\sqrt{176}}{\sqrt{6}} = 2 \sqrt{\frac{22}{3}}$

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