The shortest distance between the lines frac{ | vedclass

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(B) The given lines are $L_1: \vec{r} = (3\hat{i} - 15\hat{j} + 9\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 5\hat{k})$ and $L_2: \vec{r} = (-1\hat{i} +

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$4 \sqrt{3}$

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(B) The given lines are $L_1: \vec{r} = (3\hat{i} - 15\hat{j} + 9\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 5\hat{k})$ and $L_2: \vec{r} = (-1\hat{i} + 1\hat{j} + 9\hat{k}) + \mu(2\hat{i} + 1\hat{j} - 3\hat{k})$.Here,$\vec{a}_1 = 3\hat{i} - 15\hat{j} + 9\hat{k}$,$\vec{a}_2 = -\hat{i} + \hat{j} + 9\hat{k}$,$\vec{b}_1 = 2\hat{i} - 7\hat{j} + 5\hat{k}$,and $\vec{b}_2 = 2\hat{i} + \hat{j} - 3\hat{k}$.First,find $\vec{a}_2 - \vec{a}_1 = (-1 - 3)\hat{i} + (1 - (-15))\hat{j} + (9 - 9)\hat{k} = -4\hat{i} + 16\hat{j} + 0\hat{k}$.Next,find the cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -7 & 5 \ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = 16\hat{i} + 16\hat{j} + 16\hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{16^2 + 16^2 + 16^2} = 16\sqrt{3}$.The shortest distance $SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|} = \frac{|(-4)(16) + (16)(16) + (0)(16)|}{16\sqrt{3}} = \frac{|-64 + 256|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

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