The distance between the parallel lines frac{ | vedclass

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(C) The given lines are $L_1: \frac{x-2}{2} = \frac{y-4}{5} = \frac{z-1}{2}$ and $L_2: \frac{x-1}{3} = \frac{y+1}{5} = \frac{z+3}{2}$. Note: The lines

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$\sqrt{\frac{300}{37}}$ units

Vedclass · Answer

(C) The given lines are $L_1: \frac{x-2}{2} = \frac{y-4}{5} = \frac{z-1}{2}$ and $L_2: \frac{x-1}{3} = \frac{y+1}{5} = \frac{z+3}{2}$. Note: The lines are not parallel as the direction vectors are $\vec{b}_1 = 2\hat{i} + 5\hat{j} + 2\hat{k}$ and $\vec{b}_2 = 3\hat{i} + 5\hat{j} + 2\hat{k}$. Assuming the question implies the distance between skew lines: Let $\vec{a}_1 = 2\hat{i} + 4\hat{j} + \hat{k}$ and $\vec{a}_2 = \hat{i} - \hat{j} - 3\hat{k}$. $\vec{a}_2 - \vec{a}_1 = (1-2)\hat{i} + (-1-4)\hat{j} + (-3-1)\hat{k} = -\hat{i} - 5\hat{j} - 4\hat{k}$. The cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 5 & 2 \ 3 & 5 & 2 \end{vmatrix} = \hat{i}(10-10) - \hat{j}(4-6) + \hat{k}(10-15) = 2\hat{j} - 5\hat{k}$. The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{0^2 + 2^2 + (-5)^2} = \sqrt{29}$. The scalar triple product $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)| = |(-\hat{i} - 5\hat{j} - 4\hat{k}) \cdot (2\hat{j} - 5\hat{k})| = |0 - 10 + 20| = 10$. Distance $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|} = \frac{10}{\sqrt{29}}$. Given the provided solution steps in the prompt,there is a calculation error in the original problem statement regarding parallelism. Following the provided logic structure: $\vec{a}_2 - \vec{a}_1 = -\hat{i} - 5\hat{j} - 4\hat{k}$,$\vec{b} = 3\hat{i} + 5\hat{j} + 2\hat{k}$. The cross product $\vec{b} 	imes (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & 2 \ -1 & -5 & -4 \end{vmatrix} = \hat{i}(-20+10) - \hat{j}(-12+2) + \hat{k}(-15+5) = -10\hat{i} + 10\hat{j} - 10\hat{k}$. Magnitude $= \sqrt{100+100+100} = \sqrt{300}$. $|\vec{b}| = \sqrt{9+25+4} = \sqrt{38}$. Distance $= \sqrt{\frac{300}{38}} = \sqrt{\frac{150}{19}}$.

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

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