The shortest distance between the two lines, | vedclass

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(A) Line $1$ passes through $A(0, 0, 0)$ and $B(2, 0, 3)$. Direction vector $\vec{b_1} = (2-0)\hat{i} + (0-0)\hat{j} + (3-0)\hat{k} = 2\hat{i} + 3\hat

Vedclass · Accepted Answer

$\frac{24}{7}$ units

Vedclass · Answer

(A) Line $1$ passes through $A(0, 0, 0)$ and $B(2, 0, 3)$. Direction vector $\vec{b_1} = (2-0)\hat{i} + (0-0)\hat{j} + (3-0)\hat{k} = 2\hat{i} + 3\hat{k}$. Equation: $\vec{r} = 0\hat{i} + 0\hat{j} + 0\hat{k} + \lambda(2\hat{i} + 3\hat{k})$.Line $2$ passes through $C(2, 5, 0)$ and $D(0, 4, 0)$. Direction vector $\vec{b_2} = (0-2)\hat{i} + (4-5)\hat{j} + (0-0)\hat{k} = -2\hat{i} - \hat{j}$. Equation: $\vec{r} = (2\hat{i} + 5\hat{j}) + \mu(-2\hat{i} - \hat{j})$.Shortest distance $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.$\vec{a_2} - \vec{a_1} = 2\hat{i} + 5\hat{j}$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 3 \ -2 & -1 & 0 \end{vmatrix} = \hat{i}(0 - (-3)) - \hat{j}(0 - (-6)) + \hat{k}(-2 - 0) = 3\hat{i} - 6\hat{j} - 2\hat{k}$.Magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.Dot product $(2\hat{i} + 5\hat{j}) \cdot (3\hat{i} - 6\hat{j} - 2\hat{k}) = (2)(3) + (5)(-6) + (0)(-2) = 6 - 30 = -24$.Distance $d = \frac{|-24|}{7} = \frac{24}{7}$ units.

The shortest distance between the two lines, where the first line passes through $(0, 0, 0)$ and $(2, 0, 3)$ and the second line passes through $(2, 5, 0)$ and $(0, 4, 0)$, is

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