Let the lines L_1: frac{x + 1}{3} = frac{y + | vedclass

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(B) The direction vectors of lines $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$ res

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$\frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$

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(B) The direction vectors of lines $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.$A$ vector perpendicular to both $L_1$ and $L_2$ is given by the cross product $\vec{n} = \vec{v_1} 	imes \vec{v_2}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.The magnitude of this vector is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.The unit vector perpendicular to both lines is $\pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

Let the lines $L_1: \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_2: \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$. The unit vector perpendicular to both $L_1$ and $L_2$ is:

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