The shortest distance between the lines frac{ | vedclass

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(C) The given lines are $L_1: \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $L_2: \frac{x-6}{3}=\frac{y-1}{-2}=\frac{z+8}{0}$.Here,point $A(2, -1, 6)

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) The given lines are $L_1: \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $L_2: \frac{x-6}{3}=\frac{y-1}{-2}=\frac{z+8}{0}$.Here,point $A(2, -1, 6)$ lies on $L_1$ and point $B(6, 1, -8)$ lies on $L_2$.The direction vectors are $\vec{b_1} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} - 2\hat{j} + 0\hat{k}$.The vector $\vec{AB} = (6-2)\hat{i} + (1-(-1))\hat{j} + (-8-6)\hat{k} = 4\hat{i} + 2\hat{j} - 14\hat{k}$.The cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 2 \ 3 & -2 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 6) + \hat{k}(-6 - 6) = 4\hat{i} + 6\hat{j} - 12\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|} = \frac{|(4)(4) + (2)(6) + (-14)(-12)|}{14} = \frac{|16 + 12 + 168|}{14} = \frac{196}{14} = 14$.

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$

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