The shortest distance between the lines frac{ | vedclass

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(B) The given lines are $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$.For the first line, a point $\ve

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$4$

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(B) The given lines are $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$.For the first line, a point $\vec{a} = \hat{i} - 8\hat{j} + 4\hat{k}$ and direction vector $\vec{p} = 2\hat{i} - 7\hat{j} + 5\hat{k}$.For the second line, a point $\vec{b} = \hat{i} + 2\hat{j} + 6\hat{k}$ and direction vector $\vec{q} = 2\hat{i} + \hat{j} - 3\hat{k}$.The cross product $\vec{p} 	imes \vec{q}$ is given by:$\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -7 & 5 \ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21-5) - \hat{j}(-6-10) + \hat{k}(2+14) = 16\hat{i} + 16\hat{j} + 16\hat{k} = 16(\hat{i} + \hat{j} + \hat{k})$.The magnitude is $|\vec{p} 	imes \vec{q}| = 16\sqrt{1^2 + 1^2 + 1^2} = 16\sqrt{3}$.The vector $(\vec{a} - \vec{b}) = (1-1)\hat{i} + (-8-2)\hat{j} + (4-6)\hat{k} = -10\hat{j} - 2\hat{k}$.The shortest distance $d$ is given by $d = \left| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|} ight|$.$d = \left| \frac{(-10\hat{j} - 2\hat{k}) \cdot 16(\hat{i} + \hat{j} + \hat{k})}{16\sqrt{3}} ight| = \left| \frac{16(0 - 10 - 2)}{16\sqrt{3}} ight| = \left| \frac{-12}{\sqrt{3}} ight| = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

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