A line L is parallel to both the planes 2x + | vedclass

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(A) The line $L$ is parallel to both planes, so it is perpendicular to the normal vectors of both planes. Let the normal vectors be $\vec{n_1} = 2\hat

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

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(A) The line $L$ is parallel to both planes, so it is perpendicular to the normal vectors of both planes. Let the normal vectors be $\vec{n_1} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.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 $X$-axis is the angle between $\vec{u}$ and the unit vector $\hat{i} = (1, 0, 0)$.$\cos \alpha = \frac{\vec{u} \cdot \hat{i}}{|\vec{u}| |\hat{i}|} = \frac{(1)(1) + (-1)(0) + (1)(0)}{\sqrt{1^2 + (-1)^2 + 1^2} \cdot 1} = \frac{1}{\sqrt{3}}$.

$A$ line $L$ is parallel to both the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive direction of the $X$-axis, then $\cos \alpha =$

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