The shortest distance between the lines frac{ | vedclass

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(B) The shortest distance between two parallel lines $\vec{r} = \vec{a_1} + \lambda \vec{b}$ and $\vec{r} = \vec{a_2} + \mu \vec{b}$ is given by $d =

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$\sqrt{\frac{293}{49}}$

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(B) The shortest distance between two parallel lines $\vec{r} = \vec{a_1} + \lambda \vec{b}$ and $\vec{r} = \vec{a_2} + \mu \vec{b}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}|}{|\vec{b}|}$.Here,the lines are parallel because their direction vectors are the same,$\vec{b} = (2, 3, 6)$.The points on the lines are $\vec{a_1} = (1, 2, -4)$ and $\vec{a_2} = (3, 3, -5)$.Then,$\vec{a_2} - \vec{a_1} = (3-1, 3-2, -5-(-4)) = (2, 1, -1)$.Now,calculate the cross product $(\vec{a_2} - \vec{a_1}) 	imes \vec{b}$:$(\vec{a_2} - \vec{a_1}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 2 & 3 & 6 \end{vmatrix} = \hat{i}(6 - (-3)) - \hat{j}(12 - (-2)) + \hat{k}(6 - 2) = 9\hat{i} - 14\hat{j} + 4\hat{k}$.The magnitude of the cross product is $|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$.The magnitude of the direction vector is $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.Therefore,the shortest distance is $d = \frac{\sqrt{293}}{7} = \sqrt{\frac{293}{49}}$.

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is . . . . . . .

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