The shortest distance between the lines frac{ | vedclass

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(D) The lines are given by $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.Comparing these with $\frac{x-x

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$\frac{1}{\sqrt{6}}$ units.

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(D) The lines are given by $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.Comparing these with $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$,we get:$(x_1, y_1, z_1) = (1, 2, 3)$ and $(x_2, y_2, z_2) = (2, 4, 5)$.Direction ratios are $(a_1, b_1, c_1) = (2, 3, 4)$ and $(a_2, b_2, c_2) = (3, 4, 5)$.The shortest distance $d$ is given by the formula:$d = \frac{|\det(A)|}{\sqrt{(a_1b_2-a_2b_1)^2 + (b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2}}$where $A = \begin{bmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \ 2 & 3 & 4 \ 3 & 4 & 5 \end{bmatrix}$.Calculating the determinant: $\det(A) = 1(15-16) - 2(10-12) + 2(8-9) = 1(-1) - 2(-2) + 2(-1) = -1 + 4 - 2 = 1$.Now,calculate the denominator:$a_1b_2-a_2b_1 = (2)(4)-(3)(3) = 8-9 = -1$.$b_1c_2-b_2c_1 = (3)(5)-(4)(4) = 15-16 = -1$.$c_1a_2-c_2a_1 = (4)(3)-(5)(2) = 12-10 = 2$.Denominator $= \sqrt{(-1)^2 + (-1)^2 + (2)^2} = \sqrt{1+1+4} = \sqrt{6}$.Therefore,$d = \frac{1}{\sqrt{6}}$ units.

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

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