The shortest distance between the lines r=(-2 | vedclass

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(C) The given lines are $r = a_1 + r b_1$ and $r = a_2 + k b_2$.Here,$a_1 = -2 \hat{i} + \hat{j} - \hat{k}$,$b_1 = 2 \hat{i} + 3 \hat{j} - \hat{k}$.An

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$\frac{11}{\sqrt{6}}$

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(C) The given lines are $r = a_1 + r b_1$ and $r = a_2 + k b_2$.Here,$a_1 = -2 \hat{i} + \hat{j} - \hat{k}$,$b_1 = 2 \hat{i} + 3 \hat{j} - \hat{k}$.And $a_2 = \hat{i} - \hat{j} + 2 \hat{k}$,$b_2 = -\hat{i} + 2 \hat{j} + 4 \hat{k}$.First,calculate $a_2 - a_1 = (1 - (-2)) \hat{i} + (-1 - 1) \hat{j} + (2 - (-1)) \hat{k} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$.Next,calculate the cross product $b_1 	imes b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ -1 & 2 & 4 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(8 - 1) + \hat{k}(4 - (-3)) = 14 \hat{i} - 7 \hat{j} + 7 \hat{k}$.The magnitude $|b_1 	imes b_2| = \sqrt{14^2 + (-7)^2 + 7^2} = \sqrt{196 + 49 + 49} = \sqrt{294} = 7 \sqrt{6}$.The shortest distance is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 	imes b_2)|}{|b_1 	imes b_2|}$.$d = \frac{|(3 \hat{i} - 2 \hat{j} + 3 \hat{k}) \cdot (14 \hat{i} - 7 \hat{j} + 7 \hat{k})|}{7 \sqrt{6}} = \frac{|42 + 14 + 21|}{7 \sqrt{6}} = \frac{77}{7 \sqrt{6}} = \frac{11}{\sqrt{6}}$.

The shortest distance between the lines $r=(-2 \hat{i}+\hat{j}-\hat{k})+r(2 \hat{i}+3 \hat{j}-\hat{k})$ and $r=(\hat{i}-\hat{j}+2 \hat{k})+k(-\hat{i}+2 \hat{j}+4 \hat{k})$ is

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