The perimeter of a square whose two sides lie | vedclass

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(B) The given lines are $L_1: \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$.Since the direction ratios

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$\frac{4 \sqrt{673}}{\sqrt{29}}$ units

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(B) The given lines are $L_1: \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$.Since the direction ratios $(2, 3, 4)$ are the same,the lines are parallel.The distance $d$ between two parallel lines $\vec{r} = \vec{a_1} + t\vec{b}$ and $\vec{r} = \vec{a_2} + s\vec{b}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}|}{|\vec{b}|}$.Here,$\vec{a_1} = (1, -2, 3)$,$\vec{a_2} = (0, 1, -1)$,and $\vec{b} = (2, 3, 4)$.$\vec{a_2} - \vec{a_1} = (-1, 3, -4)$.$(\vec{a_2} - \vec{a_1}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 3 & -4 \ 2 & 3 & 4 \end{vmatrix} = \hat{i}(12 - (-12)) - \hat{j}(-4 - (-8)) + \hat{k}(-3 - 6) = 24\hat{i} - 4\hat{j} - 9\hat{k}$.The magnitude is $\sqrt{24^2 + (-4)^2 + (-9)^2} = \sqrt{576 + 16 + 81} = \sqrt{673}$.The magnitude of $\vec{b}$ is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.Thus,the side length of the square is $s = \frac{\sqrt{673}}{\sqrt{29}}$.The perimeter of the square is $4s = \frac{4\sqrt{673}}{\sqrt{29}}$ units.

The perimeter of a square whose two sides lie along the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

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