The shortest distance between the lines frac{ | vedclass

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(A) The given lines are in the form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$.Here,$\vec{a_1} = \hat{i} + 2\

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$\frac{1}{\sqrt{6}}$

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(A) The given lines are in the form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$.Here,$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Also,$\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{b_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.First,calculate the cross product $\vec{b_1} 	imes \vec{b_2}$:$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.Next,calculate $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})$:$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1$.The shortest distance $d$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.Therefore,$d = \frac{1}{\sqrt{6}}$.

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is:

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