Let L be the line of intersection of the plan | vedclass

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(C) The normal vectors to the given planes are $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + 3\hat{j} + 2\hat{k}$.Since the l

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$\frac{1}{\sqrt{3}}$

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(C) The normal vectors to the given planes are $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + 3\hat{j} + 2\hat{k}$.Since the line $L$ is the intersection of these two planes,it is perpendicular to both normal vectors.Therefore,the direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n}_1 	imes \vec{n}_2$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.We can simplify the direction vector to $\vec{u} = \hat{i} - \hat{j} + \hat{k}$.The angle $\alpha$ that the line makes with the positive $X$-axis (represented by the unit vector $\hat{i}$) is given by $\cos \alpha = \frac{\vec{u} \cdot \hat{i}}{|\vec{u}| |\hat{i}|}$.Calculating the dot product: $\vec{u} \cdot \hat{i} = (1)(1) + (-1)(0) + (1)(0) = 1$.Calculating the magnitude: $|\vec{u}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.Thus,$\cos \alpha = \frac{1}{\sqrt{3} 	imes 1} = \frac{1}{\sqrt{3}}$.

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive $X$-axis,then $\cos \alpha$ equals

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