The shortest distance between the lines frac{ | vedclass

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(C) The given lines are $L_1: \frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $L_2: \frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For $L_1$,a po

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$3\sqrt{5}$

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(C) The given lines are $L_1: \frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $L_2: \frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.For $L_1$,a point is $A(4, 3, 2)$ and the direction vector is $\vec{v}_1 = (1, 2, -3)$.For $L_2$,a point is $B(-2, 6, 5)$ and the direction vector is $\vec{v}_2 = (2, 4, -5)$.The vector $\vec{AB} = (-2-4, 6-3, 5-2) = (-6, 3, 3)$.The shortest distance $d$ between two skew lines is given by $d = \frac{|\vec{AB} \cdot (\vec{v}_1 	imes \vec{v}_2)|}{|\vec{v}_1 	imes \vec{v}_2|}$.First,calculate the cross product $\vec{v}_1 	imes \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$.The magnitude is $|\vec{v}_1 	imes \vec{v}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4+1} = \sqrt{5}$.The dot product is $\vec{AB} \cdot (\vec{v}_1 	imes \vec{v}_2) = (-6, 3, 3) \cdot (2, -1, 0) = (-6)(2) + (3)(-1) + (3)(0) = -12 - 3 + 0 = -15$.Thus,$d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.

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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