The shortest distance between the lines x−$$23=y$$−$$34=z$$−$$45$$ ;x−$$12=y$$−$$23=z$$−$$34is$$

a¯$$1=(2$$,$$3$$,$$4)$$, a¯$$2=(1$$,$$2$$,$$3)$$, b¯$$1=(3$$,$$4$$,$$5)$$, b¯$$2=(2$$,$$3$$,$$4)$$, a¯$$1$$−a¯$$2$$ b¯$$1$$ b¯$$2=0$$

Straight lines in 3D

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Question

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}; \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is

Moderate

Solution

${\bar{a}}_{1} = (2, 3, 4)$ , ${\bar{a}}_{2} = (1, 2, 3)$ , ${\bar{b}}_{1} = (3, 4, 5)$ , ${\bar{b}}_{2} = (2, 3, 4)$ , $[{\bar{a}}_{1} - {\bar{a}}_{2} {\bar{b}}_{1} {\bar{b}}_{2}] = 0$

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