The shortest distance between lines L_1 and L | vedclass

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(C) First,find the equation of line $L_2$. The direction vector of $L_2$ is perpendicular to both the vector $\vec{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (

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$\frac{141}{\sqrt{221}}$

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(C) First,find the equation of line $L_2$. The direction vector of $L_2$ is perpendicular to both the vector $\vec{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$ and the direction vector of the given line $\vec{v} = (-2, 3, 1)$.Thus,the direction vector $\vec{n_2}$ of $L_2$ is $\vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 0 \ -2 & 3 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(3-0) + \hat{k}(9-(-4)) = 2\hat{i} - 3\hat{j} + 13\hat{k}$.However,the problem states $L_2$ passes through $A(-4, 4, 3)$ and $B(-1, 6, 3)$,so its direction vector is simply $\vec{AB} = (3, 2, 0)$.The shortest distance $d$ between $L_1: \vec{r} = (1, -1, -4) + \lambda(2, -3, 2)$ and $L_2: \vec{r} = (-4, 4, 3) + \mu(3, 2, 0)$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} 	imes \vec{n_2})|}{ |\vec{n_1} 	imes \vec{n_2}| }$.$\vec{a_2} - \vec{a_1} = (-4-1, 4-(-1), 3-(-4)) = (-5, 5, 7)$.$\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 2 \ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0-6) + \hat{k}(4-(-9)) = -4\hat{i} + 6\hat{j} + 13\hat{k}$.$|\vec{n_1} 	imes \vec{n_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$.$|(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} 	imes \vec{n_2})| = |(-5)(-4) + (5)(6) + (7)(13)| = |20 + 30 + 91| = 141$.Therefore,$d = \frac{141}{\sqrt{221}}$.

The shortest distance between lines $L_1$ and $L_2$,where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$ and $B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$,is

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