The shortest distance between the lines frac{ | vedclass

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(B) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2}

Vedclass · Accepted Answer

$(2, 3]$

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(B) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ is given by the formula:$d = \frac{|(x_2-x_1, y_2-y_1, z_2-z_1) \cdot (a_1, b_1, c_1) 	imes (a_2, b_2, c_2)|}{|(a_1, b_1, c_1) 	imes (a_2, b_2, c_2)|}$For the given lines,we have $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a_1, b_1, c_1) = (2, 2, 1)$.For the second line,$(x_2, y_2, z_2) = (-2, 4, 5)$ and $(a_2, b_2, c_2) = (-1, 8, 4)$.The vector product $(a_1, b_1, c_1) 	imes (a_2, b_2, c_2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 1 \ -1 & 8 & 4 \end{vmatrix} = \hat{i}(8-8) - \hat{j}(8+1) + \hat{k}(16+2) = (0, -9, 18)$.The magnitude is $\sqrt{0^2 + (-9)^2 + 18^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$.The scalar triple product is $\begin{vmatrix} -2 & 4 & 5 \ 2 & 2 & 1 \ -1 & 8 & 4 \end{vmatrix} = -2(8-8) - 4(8+1) + 5(16+2) = 0 - 36 + 90 = 54$.Thus,$d = \frac{|54|}{9\sqrt{5}} = \frac{6}{\sqrt{5}} = \frac{6 	imes 2.236}{5} \approx 2.68$.Since $2.68 \in (2, 3]$,the correct option is $B$.

The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

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