The shortest distance between the lines frac{ | vedclass

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(A) The shortest distance $S_d$ between two skew lines $\vec{r} = \vec{a} + \lambda \vec{n}_1$ and $\vec{r} = \vec{b} + \mu \vec{n}_2$ is given by the

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$3 \sqrt{6}$

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(A) The shortest distance $S_d$ between two skew lines $\vec{r} = \vec{a} + \lambda \vec{n}_1$ and $\vec{r} = \vec{b} + \mu \vec{n}_2$ is given by the formula: $S_d = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{n}_1 	imes \vec{n}_2)}{|\vec{n}_1 	imes \vec{n}_2|} ight|$ From the given equations: $\vec{a} = (4, -2, -3)$,$\vec{b} = (1, 3, 4)$ $\vec{n}_1 = (4, 5, 3)$,$\vec{n}_2 = (3, 4, 2)$ First,calculate the cross product $\vec{n}_1 	imes \vec{n}_2$: $\vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 5 & 3 \ 3 & 4 & 2 \end{vmatrix} = \hat{i}(10-12) - \hat{j}(8-9) + \hat{k}(16-15) = -2\hat{i} + \hat{j} + \hat{k} = (-2, 1, 1)$ The magnitude is $|\vec{n}_1 	imes \vec{n}_2| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$ Now,calculate $\vec{b} - \vec{a} = (1-4, 3-(-2), 4-(-3)) = (-3, 5, 7)$ The dot product $(\vec{b} - \vec{a}) \cdot (\vec{n}_1 	imes \vec{n}_2) = (-3, 5, 7) \cdot (-2, 1, 1) = 6 + 5 + 7 = 18$ Therefore,$S_d = \left| \frac{18}{\sqrt{6}} ight| = \frac{18}{\sqrt{6}} 	imes \frac{\sqrt{6}}{\sqrt{6}} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}$

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

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