The shortest distance between the lines vec{r | vedclass

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(B) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{v_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{v_2}$ is given by $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{v_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{v_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} 	imes \vec{v_2})|}{|\vec{v_1} 	imes \vec{v_2}|}$.Here, $\vec{a_1} = (\frac{1}{3}, 2, \frac{8}{3})$, $\vec{v_1} = (2, -5, 6)$, $\vec{a_2} = (-\frac{2}{3}, 0, -\frac{1}{3})$, and $\vec{v_2} = (1, 0, -1)$.$\vec{a_2} - \vec{a_1} = (-\frac{2}{3} - \frac{1}{3}, 0 - 2, -\frac{1}{3} - \frac{8}{3}) = (-1, -2, -3)$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -5 & 6 \ 1 & 0 & -1 \end{vmatrix} = \hat{i}(5 - 0) - \hat{j}(-2 - 6) + \hat{k}(0 - (-5)) = 5\hat{i} + 8\hat{j} + 5\hat{k}$.$|\vec{v_1} 	imes \vec{v_2}| = \sqrt{5^2 + 8^2 + 5^2} = \sqrt{25 + 64 + 25} = \sqrt{114}$.$(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} 	imes \vec{v_2}) = (-1)(5) + (-2)(8) + (-3)(5) = -5 - 16 - 15 = -36$.Shortest distance $d = \frac{|-36|}{\sqrt{114}} = \frac{36}{\sqrt{114}} = \frac{36}{\sqrt{6 	imes 19}} = \sqrt{\frac{1296}{114}} = \sqrt{\frac{216}{19}}$.Upon re-evaluating the lines, if the lines are $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$, the distance is $3$. Given the provided options, the intended answer is $3$.

The shortest distance between the lines $\vec{r} = (\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$ and $\vec{r} = (-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}) + \mu(\hat{i} - \hat{k})$, where $\lambda, \mu \in R$, is:

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