The shortest distance between the skew lines | vedclass

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Question

(A) 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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$\frac{3 \sqrt{2}}{\sqrt{7}}$

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(A) 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 $d = \left| \frac{(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 	imes \vec{b}_2|} ight|$.Given $\vec{a}_1 = 2\hat{i} - \hat{j}$,$\vec{b}_1 = \hat{i} + 2\hat{k}$ and $\vec{a}_2 = -2\hat{i} + \hat{k}$,$\vec{b}_2 = \hat{i} - \hat{j} - \hat{k}$.First,calculate the cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 2 \ 1 & -1 & -1 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(-1 - 2) + \hat{k}(-1 - 0) = 2\hat{i} + 3\hat{j} - \hat{k}$.The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.Next,calculate $\vec{a}_2 - \vec{a}_1 = (-2\hat{i} + \hat{k}) - (2\hat{i} - \hat{j}) = -4\hat{i} + \hat{j} + \hat{k}$.Now,calculate the dot product $(\vec{b}_1 	imes \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (2\hat{i} + 3\hat{j} - \hat{k}) \cdot (-4\hat{i} + \hat{j} + \hat{k}) = (2)(-4) + (3)(1) + (-1)(1) = -8 + 3 - 1 = -6$.The shortest distance is $d = \left| \frac{-6}{\sqrt{14}} ight| = \frac{6}{\sqrt{14}} = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7} = \frac{3\sqrt{2}\sqrt{7}}{7} = \frac{3\sqrt{2}}{\sqrt{7}}$.

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

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