The shortest distance between the lines frac{ | vedclass

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(C) The lines are given by $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For line $1$, a point

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$3\sqrt{5}$

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(C) The lines are given by $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For line $1$, a point $P_1 = (4, 3, 2)$ and direction vector $\vec{v}_1 = (1, 2, -3)$.For line $2$, a point $P_2 = (-2, 6, 5)$ and direction vector $\vec{v}_2 = (2, 4, -5)$.The vector connecting the two points is $\vec{P_1P_2} = (-2-4, 6-3, 5-2) = (-6, 3, 3)$.The cross product of the direction vectors is $\vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$.The shortest distance $d$ is given by $d = \frac{|\vec{P_1P_2} \cdot (\vec{v}_1 	imes \vec{v}_2)|}{|\vec{v}_1 	imes \vec{v}_2|}$.Calculating the dot product: $|(-6, 3, 3) \cdot (2, -1, 0)| = |-12 - 3 + 0| = |-15| = 15$.Calculating the magnitude of the cross product: $|\vec{v}_1 	imes \vec{v}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5}$.Therefore, $d = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.

The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ is:

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