Statement 1: The shortest distance between th | vedclass

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(B) The given lines are $L_1: \frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $L_2: \frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$.Both lines have the

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true,Statement $2$ is a correct explanation for Statement $1$.

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(B) The given lines are $L_1: \frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $L_2: \frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$.Both lines have the same direction vector $\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}$,so they are parallel.The formula for the shortest distance between two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is $d = \frac{|(\vec{a}_2 - \vec{a}_1) 	imes \vec{b}|}{|\vec{b}|}$.Here,$\vec{a}_1 = 0\hat{i} + 0\hat{j} + 0\hat{k}$ and $\vec{a}_2 = 1\hat{i} + 1\hat{j} + 1\hat{k}$.Thus,$\vec{a}_2 - \vec{a}_1 = \hat{i} + \hat{j} + \hat{k}$.$(\vec{a}_2 - \vec{a}_1) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(2 - 2) + \hat{k}(-1 - 2) = 3\hat{i} + 0\hat{j} - 3\hat{k}$.The magnitude is $|3\hat{i} - 3\hat{k}| = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.The magnitude of $\vec{b}$ is $|2\hat{i} - \hat{j} + 2\hat{k}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$.Therefore,$d = \frac{3\sqrt{2}}{3} = \sqrt{2}$.Since the calculated distance is $\sqrt{2}$,Statement $1$ is true.Statement $2$ is the standard definition of the shortest distance between parallel lines,which is also true.Since Statement $2$ provides the method used to calculate the distance in Statement $1$,it is the correct explanation.

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