If the square of the shortest distance betwee | vedclass

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(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{p}$ and $\vec{r} = \vec{a_2} + \mu \vec{q}$,where $\vec{a_1} = (2, 1, -3)$,$\vec{p} = (

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$9$

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(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{p}$ and $\vec{r} = \vec{a_2} + \mu \vec{q}$,where $\vec{a_1} = (2, 1, -3)$,$\vec{p} = (1, 2, -3)$,$\vec{a_2} = (-1, -3, -5)$,and $\vec{q} = (2, 4, -5)$.The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} 	imes \vec{q})|}{|\vec{p} 	imes \vec{q}|}$.First,calculate $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j} + 0\hat{k}$.The magnitude is $|\vec{p} 	imes \vec{q}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.Next,$\vec{a_2} - \vec{a_1} = (-1-2, -3-1, -5-(-3)) = (-3, -4, -2)$.The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} 	imes \vec{q}) = (-3, -4, -2) \cdot (2, -1, 0) = -6 + 4 + 0 = -2$.Thus,$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.The square of the shortest distance is $d^2 = \frac{4}{5}$.Given $d^2 = \frac{m}{n} = \frac{4}{5}$,where $m=4$ and $n=5$ are coprime.Therefore,$m+n = 4+5 = 9$.

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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