The shortest distance between the lines r→$$=a$$→$$1+b$$→$$1$$ t and r→$$=a$$→$$2+b$$→$$2$$ s is

Correct option is 2[a→2−a→1 , b→1 , b→2]|b→1 × b→2|

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The shortest distance between the lines $\vec{r} = {\vec{a}}_{1} + {\vec{b}}_{1} t$ and $\vec{r} = {\vec{a}}_{2} + {\vec{b}}_{2} s$ is

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[a→2−a→1 , b→1 , b→2]

[a→2−a→1 , b→1 , b→2]|b→1 × b→2|

[a→2−a→1 , b→1 , b→2]|a→1 × a→2|

none of these

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detailed solution

Correct option is B

Standard formula for shortest distance.

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The shortest distance between the lines r→=a→1+b→1 t and r→=a→2+b→2 s is

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The shortest distance between the lines $\vec{r} = {\vec{a}}_{1} + {\vec{b}}_{1} t$ and $\vec{r} = {\vec{a}}_{2} + {\vec{b}}_{2} s$ is