The shortest distance between the lines frac{ | vedclass

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(A) The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \

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$\frac{2}{\sqrt{69}}$ units

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(A) The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$. For the given lines: Line $1$: $\vec{a_1} = (-1, 2, -1)$,$\vec{b_1} = (3, 2, 2)$. Line $2$: $\vec{a_2} = (2, 2, -3)$,$\vec{b_2} = (1, 2, 3)$. $\vec{a_2} - \vec{a_1} = (2 - (-1), 2 - 2, -3 - (-1)) = (3, 0, -2)$. $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(6 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 2) = 2\hat{i} - 7\hat{j} + 4\hat{k}$. $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + (-7)^2 + 4^2} = \sqrt{4 + 49 + 16} = \sqrt{69}$. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (3)(2) + (0)(-7) + (-2)(4) = 6 + 0 - 8 = -2$. $d = \frac{|-2|}{\sqrt{69}} = \frac{2}{\sqrt{69}}$ units.

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

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