First slide

Straight lines in 3D

Question

The shortest distance between the lines
$\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} and \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} is:$

Difficult

A

$\frac{7}{2} \sqrt{30}$
B

$2 \sqrt{30}$
C

3
D

$3 \sqrt{30}$

Solution

$\begin{array}{l} \vec{A B} = 6 \hat{i} + 15 \hat{j} + 3 \hat{k} \\ \bar{p} = \hat{i} + 4 \hat{j} + 22 \hat{k} \\ \bar{q} = \hat{i} + \hat{j} + 7 \hat{k} \\ \bar{p} \times \bar{q} = |\begin{array}{l} i & j & k \\ 1 & 4 & 22 \\ 1 & 1 & 7 \end{array}| = 6 \hat{i} + 15 \hat{j} - 3 \hat{k} \\ S.D = \frac{| A B \cdot (\bar{p} \times \bar{q}) |}{| \bar{p} \times \vec{q} |} = \frac{| 36 + 225 + 9 |}{\sqrt{36 + 225 + 9}} = 3 \sqrt{30} \end{array}$

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