The shortest distance between the lines overl | vedclass

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Question

(B) The shortest distance $d$ between two lines $\overline{r} = \overline{a_1} + t\overline{b_1}$ and $\overline{r} = \overline{a_2} + s\overline{b_2}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The shortest distance $d$ between two lines $\overline{r} = \overline{a_1} + t\overline{b_1}$ and $\overline{r} = \overline{a_2} + s\overline{b_2}$ is given by the formula $d = \frac{|(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} 	imes \overline{b_2})|}{||\overline{b_1} 	imes \overline{b_2}||}$.Here, $\overline{a_1} = 3\bar{i} - 5\bar{j} + 2\bar{k}$, $\overline{b_1} = 4\bar{i} + 3\bar{j} - \bar{k}$, $\overline{a_2} = \bar{i} + 2\bar{j} - 4\bar{k}$, and $\overline{b_2} = 6\bar{i} + 3\bar{j} - 2\bar{k}$.First, calculate $\overline{a_2} - \overline{a_1} = (1-3)\bar{i} + (2-(-5))\bar{j} + (-4-2)\bar{k} = -2\bar{i} + 7\bar{j} - 6\bar{k}$.Next, calculate the cross product $\overline{b_1} 	imes \overline{b_2} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 4 & 3 & -1 \ 6 & 3 & -2 \end{vmatrix} = \bar{i}(-6 - (-3)) - \bar{j}(-8 - (-6)) + \bar{k}(12 - 18) = -3\bar{i} + 2\bar{j} - 6\bar{k}$.The magnitude $||\overline{b_1} 	imes \overline{b_2}|| = \sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.The dot product $(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} 	imes \overline{b_2}) = (-2)(-3) + (7)(2) + (-6)(-6) = 6 + 14 + 36 = 56$.Finally, $d = \frac{|56|}{7} = 8$.

The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

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