The shortest distance between the lines 1+x=2 | vedclass

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(C) First,we write the equations of the lines in symmetric form $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$.For the first line $1+x =

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$2$ units

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(C) First,we write the equations of the lines in symmetric form $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$.For the first line $1+x = 2y = -12z$,we have $\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$. Here,point $P_1 = (-1, 0, 0)$ and direction vector $\vec{b_1} = (1, 1/2, -1/12)$.For the second line $x = y+2 = 6z-6$,we have $\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$. Here,point $P_2 = (0, -2, 1)$ and direction vector $\vec{b_2} = (1, 1, 1/6)$.The shortest distance $d$ is given by $d = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{||\vec{b_1} 	imes \vec{b_2}||}$.Vector $\vec{P_2} - \vec{P_1} = (0 - (-1), -2 - 0, 1 - 0) = (1, -2, 1)$.Cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1/2 & -1/12 \ 1 & 1 & 1/6 \end{vmatrix} = \hat{i}(\frac{1}{12} + \frac{1}{12}) - \hat{j}(\frac{1}{6} + \frac{1}{12}) + \hat{k}(1 - 1/2) = (1/6, -1/4, 1/2)$.Magnitude $||\vec{b_1} 	imes \vec{b_2}|| = \sqrt{(1/6)^2 + (-1/4)^2 + (1/2)^2} = \sqrt{1/36 + 1/16 + 1/4} = \sqrt{\frac{4+9+36}{144}} = \sqrt{49/144} = 7/12$.Dot product $(\vec{P_2} - \vec{P_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (1)(1/6) + (-2)(-1/4) + (1)(1/2) = 1/6 + 1/2 + 1/2 = 7/6$.Therefore,$d = \frac{|7/6|}{7/12} = \frac{7}{6} 	imes \frac{12}{7} = 2$ units.

The shortest distance between the lines $1+x=2y=-12z$ and $x=y+2=6z-6$ is

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