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(A) The line $L$ is the intersection of two planes: $2x + 3y + z = 1$ and $x + 3y + 2z = 2$.The direction vector $\vec{v}$ of the line $L$ is given by

Vedclass · Accepted Answer

$\sqrt{3}$

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(A) The line $L$ is the intersection of two planes: $2x + 3y + z = 1$ and $x + 3y + 2z = 2$.The direction vector $\vec{v}$ of the line $L$ is given by the cross product of the normals to the planes,$\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.$\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.We can simplify the direction vector to $\vec{v} = (1, -1, 1)$.The line $L$ makes an angle $\alpha$ with the positive $X$-axis,which has the direction vector $\vec{u} = (1, 0, 0)$.The cosine of the angle $\alpha$ is given by $\cos \alpha = \frac{|\vec{v} \cdot \vec{u}|}{|\vec{v}| |\vec{u}|}$.$\cos \alpha = \frac{|(1)(1) + (-1)(0) + (1)(0)|}{\sqrt{1^2 + (-1)^2 + 1^2} \cdot \sqrt{1^2 + 0^2 + 0^2}} = \frac{1}{\sqrt{3} \cdot 1} = \frac{1}{\sqrt{3}}$.Therefore,$\sec \alpha = \frac{1}{\cos \alpha} = \sqrt{3}$.

If a line $L$ is the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive $X$-axis,then the value of $\sec \alpha$ is

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