The shortest distance between the lines frac{ | vedclass

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(D) The shortest distance between two lines $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ and $\frac{x - x_2}{a_2} = \frac{y - y_2

Vedclass · Accepted Answer

$3\sqrt{30}$

Vedclass · Answer

(D) The shortest distance between two lines $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ and $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$ is given by the formula: $d = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.Here,$\vec{r_1} = (3, 8, 3)$ and $\vec{r_2} = (-3, -7, 6)$,so $\vec{r_2} - \vec{r_1} = (-6, -15, 3)$.The direction vectors are $\vec{b_1} = (3, -1, 1)$ and $\vec{b_2} = (-3, 2, 4)$.Calculate the cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4 - 2) - \hat{j}(12 + 3) + \hat{k}(6 - 3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.The dot product $(\vec{r_2} - \vec{r_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.Thus,$d = \frac{|270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30}$.

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

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