The shortest distance between the lines frac{ | vedclass

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(C) The lines are given by $\vec{r} = \vec{a} + \lambda \vec{p}$ and $\vec{r} = \vec{b} + \mu \vec{q}$.Here,$\vec{a} = 2\hat{i} + 4\hat{j} + 5\hat{k}$

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$\frac{1}{\sqrt{6}}$

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(C) The lines are given by $\vec{r} = \vec{a} + \lambda \vec{p}$ and $\vec{r} = \vec{b} + \mu \vec{q}$.Here,$\vec{a} = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{p} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.Also,$\vec{b} = 1\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{q} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.The shortest distance $SD$ is given by the formula $SD = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|} ight|$.First,calculate $\vec{b} - \vec{a} = (1-2)\hat{i} + (2-4)\hat{j} + (3-5)\hat{k} = -\hat{i} - 2\hat{j} - 2\hat{k}$.Next,calculate $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$.The magnitude $|\vec{p} 	imes \vec{q}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.The dot product $(\vec{b} - \vec{a}) \cdot (\vec{p} 	imes \vec{q}) = (-1)(1) + (-2)(-2) + (-2)(1) = -1 + 4 - 2 = 1$.Thus,$SD = \left| \frac{1}{\sqrt{6}} ight| = \frac{1}{\sqrt{6}}$.

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

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