The shortest distance between the skew lines | vedclass

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Question

(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by

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$\sqrt{3}$

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(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$.Given $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b_1} = \hat{i} + 3\hat{j} + 2\hat{k}$, $\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$, and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.First, calculate $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.Next, calculate the cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & 2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(1-4) + \hat{k}(3-6) = -3\hat{i} + 3\hat{j} - 3\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-3)^2 + 3^2 + (-3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$.Now, the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-3\hat{i} + 3\hat{j} - 3\hat{k}) = -9 + 9 - 9 = -9$.The shortest distance $d = \frac{|-9|}{3\sqrt{3}} = \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

The shortest distance between the skew lines $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(\hat{i}+3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+s(2 \hat{i}+3 \hat{j}+\hat{k})$ is

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