The shortest distance between the lines overl | vedclass

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(B) The shortest distance $d$ between two lines $\overline{r} = \overline{a_1} + \lambda\overline{b_1}$ and $\overline{r} = \overline{a_2} + \mu\overl

Vedclass · Accepted Answer

$\frac{6}{\sqrt{5}}$ units

Vedclass · Answer

(B) The shortest distance $d$ between two lines $\overline{r} = \overline{a_1} + \lambda\overline{b_1}$ and $\overline{r} = \overline{a_2} + \mu\overline{b_2}$ is given by the formula $d = \frac{|(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} 	imes \overline{b_2})|}{ |\overline{b_1} 	imes \overline{b_2}| }$.Given $\overline{a_1} = 4\hat{i} - \hat{j}$,$\overline{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$,$\overline{a_2} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overline{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$.First,calculate $\overline{a_2} - \overline{a_1} = (1-4)\hat{i} + (-1 - (-1))\hat{j} + (2-0)\hat{k} = -3\hat{i} + 2\hat{k}$.Next,calculate the cross product $\overline{b_1} 	imes \overline{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 - (-12)) - \hat{j}(-5 - (-6)) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j} + 0\hat{k}$.The magnitude $|\overline{b_1} 	imes \overline{b_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}$.Now,calculate the dot product $(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} 	imes \overline{b_2}) = (-3\hat{i} + 0\hat{j} + 2\hat{k}) \cdot (2\hat{i} - \hat{j} + 0\hat{k}) = (-3)(2) + (0)(-1) + (2)(0) = -6$.The shortest distance $d = \frac{|-6|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$ units.

The shortest distance between the lines $\overline{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\overline{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k})$ is:

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