Consider the lines L_1: frac{x+1}{3}=frac{y+2 | vedclass

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(B) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$

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

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(B) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 	imes \vec{b}_2}{|\vec{b}_1 	imes \vec{b}_2|}$.First,calculate the cross product $\vec{b}_1 	imes \vec{b}_2$:$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.Next,calculate the magnitude $|\vec{b}_1 	imes \vec{b}_2|$:$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.Thus,the unit vector is $\hat{n} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

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

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