The shortest distance between the lines frac{ | vedclass

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(B) The given lines are $L_1: \frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $L_2: \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$.From $L_1$,a point $P_1 =

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$9$

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(B) The given lines are $L_1: \frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $L_2: \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$.From $L_1$,a point $P_1 = (-2, 0, 5)$ and direction vector $\vec{b_1} = \hat{i} - 2\hat{j} + 2\hat{k}$.From $L_2$,a point $P_2 = (4, 1, -3)$ and direction vector $\vec{b_2} = \hat{i} + 2\hat{j} + 0\hat{k}$.The vector $\vec{P_1P_2} = (4 - (-2))\hat{i} + (1 - 0)\hat{j} + (-3 - 5)\hat{k} = 6\hat{i} + \hat{j} - 8\hat{k}$.The cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(2 - (-2)) = -4\hat{i} + 2\hat{j} + 4\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.The shortest distance $d = \frac{|\vec{P_1P_2} \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|} = \frac{|(6\hat{i} + \hat{j} - 8\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 4\hat{k})|}{6} = \frac{|-24 + 2 - 32|}{6} = \frac{|-54|}{6} = 9$.

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.

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