The shortest distance between the lines L_1: | vedclass

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(A) The lines are given by $L_1: \bar{r} = \bar{a} + \lambda \bar{b}$ and $L_2: \bar{r} = \bar{c} + \mu \bar{d}$.Here,$\bar{a} = \hat{i} + \hat{j}$,$\

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$\frac{1}{\sqrt{14}}$

Vedclass · Answer

(A) The lines are given by $L_1: \bar{r} = \bar{a} + \lambda \bar{b}$ and $L_2: \bar{r} = \bar{c} + \mu \bar{d}$.Here,$\bar{a} = \hat{i} + \hat{j}$,$\bar{b} = \hat{i} + \hat{j} - \hat{k}$,$\bar{c} = \hat{j} + \hat{k}$,and $\bar{d} = -\hat{i} + \hat{j} + 2\hat{k}$.First,calculate $\bar{a} - \bar{c} = (\hat{i} + \hat{j}) - (\hat{j} + \hat{k}) = \hat{i} - \hat{k}$.Next,calculate the cross product $\bar{b} 	imes \bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ -1 & 1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(2 - 1) + \hat{k}(1 - (-1)) = 3\hat{i} - \hat{j} + 2\hat{k}$.The magnitude $|\bar{b} 	imes \bar{d}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.The shortest distance $D$ is given by $D = \frac{|(\bar{a} - \bar{c}) \cdot (\bar{b} 	imes \bar{d})|}{|\bar{b} 	imes \bar{d}|}$.$D = \frac{|(\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})|}{\sqrt{14}} = \frac{|(1)(3) + (0)(-1) + (-1)(2)|}{\sqrt{14}} = \frac{|3 - 2|}{\sqrt{14}} = \frac{1}{\sqrt{14}}$.

The shortest distance between the lines $L_1: \bar{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + \hat{j} - \hat{k})$ and $L_2: \bar{r} = \hat{j} + \hat{k} + \mu(\hat{j} + 2\hat{k} - \hat{i})$ is equal to:

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