If the square of the shortest distance between the lines x−21=y−12=z+3−3 and x+12=y+34=z+5−5 is mn, where m, n are coprime numbers, then m + n is equal to:

a→=(2,1,−3)b→=(−1,−3,−5)p→×q→=i^j^k^12−324−5=2i^−j^b→−a→=−3i^−4j^−2k^Sd=|(b→−a→)⋅(p→×q→)||p→×q→|−=25Sd2=45m=4,n=5⇒m+n=9

If the square of the shortest distance between the lines x−21=y−12=z+3−3 and x+12=y+34=z+5−5 is mn, where m, n are coprime numbers, then m + n is equal to:

a→=(2,1,−3)b→=(−1,−3,−5)p→×q→=i^j^k^12−324−5=2i^−j^b→−a→=−3i^−4j^−2k^Sd=|(b→−a→)⋅(p→×q→)||p→×q→|−=25Sd2=45m=4,n=5⇒m+n=9

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If the square of the shortest distance between the lines $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3} and \frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}$ is $\frac{m}{n}$ , where m, n are coprime numbers, then m + n is equal to:

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Detailed Solution

$\begin{array}{l} \vec{a} = (2, 1, - 3) \\ \vec{b} = (- 1, - 3, - 5) \end{array}$
$\begin{array}{l} \vec{p} \times \vec{q} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{array}| \\ = 2 \hat{i} - \hat{j} \\ \vec{b} - \vec{a} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k} \\ S_{d} = \frac{| (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) |}{| \vec{p} \times \vec{q} |} - \\ = \frac{2}{\sqrt{5}} \\ {(S_{d})}^{2} = \frac{4}{5} \\ m = 4, n = 5 \Rightarrow m + n = 9 \end{array}$