Write the formula for the shortest distance between the lines $\vec{r} = \vec{a_{1}} + λ \vec{b} and \vec{r} = \vec{a_{2}} + μ \vec{b} .$

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Solution

The shortest distance d between the parallel lines $\vec{r} = \vec{a_{1}} + λ \vec{b} and \vec{r} = \vec{a_{2}} + μ \vec{b}$ is given by

$d = \frac{|(\vec{a_{2}} - \vec{a_{1}}) \times \vec{b}|}{|\vec{b}|}$

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Similar questions

¯ ¯ ¯ r = {¯ ¯ ¯ a}_{1} + λ ¯ ¯ b

¯ ¯ ¯ r = {¯ ¯ ¯ a}_{2} + μ ¯ ¯ b

r = a_{1} + λ b

r = a_{2} + μ b

\vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}

\to r =_{1} + t_{1} and \to r =_{2} + n - \to a_{2}, where_{1} = 9 j + 2 k,_{1} = 3 i - j + k,_{2} = - 6 i - 5 j + 10 k,_{2} = - 3 i + 2 j + 4 k

3 \sqrt{a}

a

\vec{r} = \vec{a} + λ \vec{b} and \vec{r} = \vec{a} + μ \vec{c} .

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