The shortest distance between the skew lines | vedclass

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Question

(C) The shortest distance $d$ between two skew 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

$\frac{31}{\sqrt{59}}$

Vedclass · Answer

(C) The shortest distance $d$ between two skew 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: $d = \frac{|(x_2-x_1, y_2-y_1, z_2-z_1) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| } = \frac{|\det \begin{bmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{bmatrix}|}{\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}}$ For the given lines: $(x_1, y_1, z_1) = (2, 3, -5)$ and $(a_1, b_1, c_1) = (1, -2, 1)$ $(x_2, y_2, z_2) = (1, -2, 4)$ and $(a_2, b_2, c_2) = (-1, 3, 2)$ Vector $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ -1 & 3 & 2 \end{vmatrix} = \hat{i}(-4-3) - \hat{j}(2+1) + \hat{k}(3-2) = -7\hat{i} - 3\hat{j} + 1\hat{k}$ Magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-7)^2 + (-3)^2 + 1^2} = \sqrt{49+9+1} = \sqrt{59}$ Vector $(x_2-x_1, y_2-y_1, z_2-z_1) = (1-2, -2-3, 4-(-5)) = (-1, -5, 9)$ Dot product $= |(-1)(-7) + (-5)(-3) + (9)(1)| = |7 + 15 + 9| = 31$ Shortest distance $d = \frac{31}{\sqrt{59}}$

The shortest distance between the skew lines $\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z+5}{1}$ and $\frac{x-1}{-1}=\frac{y+2}{3}=\frac{z-4}{2}$ is

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