The shortest distance between the skew-lines | vedclass

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Question

(D) The equations of the given lines in vector form are:$L_1: \vec{r} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k}) + \lambda(-\hat{i} + 2 \hat{j} + \hat{k})$

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$\sqrt{35}$

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(D) The equations of the given lines in vector form are:$L_1: \vec{r} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k}) + \lambda(-\hat{i} + 2 \hat{j} + \hat{k})$$L_2: \vec{r} = (\hat{i} - 7 \hat{j} - 2 \hat{k}) + \mu(\hat{i} + 3 \hat{j} + 2 \hat{k})$Here, $\vec{a}_1 = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$, $\vec{a}_2 = \hat{i} - 7 \hat{j} - 2 \hat{k}$, $\vec{b}_1 = -\hat{i} + 2 \hat{j} + \hat{k}$, and $\vec{b}_2 = \hat{i} + 3 \hat{j} + 2 \hat{k}$.First, calculate $\vec{a}_2 - \vec{a}_1 = (1-3)\hat{i} + (-7-4)\hat{j} + (-2 - (-2))\hat{k} = -2 \hat{i} - 11 \hat{j} + 0 \hat{k}$.Next, calculate the cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3 \hat{j} - 5 \hat{k}$.The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.The shortest distance $d$ is given by $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)}{|\vec{b}_1 	imes \vec{b}_2|} ight|$.$d = \left| \frac{(-2 \hat{i} - 11 \hat{j} + 0 \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 5 \hat{k})}{\sqrt{35}} ight| = \left| \frac{-2 - 33 + 0}{\sqrt{35}} ight| = \left| \frac{-35}{\sqrt{35}} ight| = \sqrt{35}$.

The shortest distance between the skew-lines $\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}$ and $\frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}$ is

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