Find-the-shortest-distance-between-the-skew-lines-x-3-3-8-y-1-z-3-1-and-x-3-3-y-7-2-z-6-4-

Question Number 98119 by bobhans last updated on 11/Jun/20

Find the shortest distance between the

skew lines \frac{x - 3}{3} = \frac{8 - y}{1} = \frac{z - 3}{1} and

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} .

Commented by john santu last updated on 11/Jun/20

shortest distance lies along a direction

which ⊥ to both lines is given

the cross product of vectors a long

given two lines, l_{1}, l_{2}

\vec{n} = | \begin{matrix} 3 - 1 1 \\ - 3 2 4 \end{matrix} |

\vec{n} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}

shortest distance

SD = A \vec{B} . \vec{n} = \frac{1}{\sqrt{30}} (12 + 75 + 3)

= \frac{90}{\sqrt{30}} = 3 \sqrt{30}