The shortest distance between the lines frac{ | vedclass

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(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$.Here, $\vec{a_1} = 3\hat{i} + 8\hat{j}

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$3\sqrt{30}$

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(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$.Here, $\vec{a_1} = 3\hat{i} + 8\hat{j} + 3\hat{k}$, $\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$.$\vec{a_2} = -3\hat{i} - 7\hat{j} + 6\hat{k}$, $\vec{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.$\vec{a_2} - \vec{a_1} = (-3-3)\hat{i} + (-7-8)\hat{j} + (6-3)\hat{k} = -6\hat{i} - 15\hat{j} + 3\hat{k}$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12+3) + \hat{k}(6-3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.Shortest distance $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|} = \frac{|(-6)(-6) + (-15)(-15) + (3)(3)|}{3\sqrt{30}} = \frac{|36 + 225 + 9|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30}$.

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

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