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

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(B) Let the direction ratios of the line $L$ be $(a, b, c)$.Since the line $L$ lies in both planes,it is perpendicular to the normals of both planes.T

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the direction ratios of the line $L$ be $(a, b, c)$.Since the line $L$ lies in both planes,it is perpendicular to the normals of both planes.The normal vectors are $\vec{n_1} = (2, 3, 1)$ and $\vec{n_2} = (1, 3, 2)$.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 $(1, -1, 1)$.The direction cosines $(\ell, m, n)$ are obtained by normalizing the vector $(1, -1, 1)$:Magnitude $= \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.Thus,$\ell = \frac{1}{\sqrt{3}}$,$m = -\frac{1}{\sqrt{3}}$,$n = \frac{1}{\sqrt{3}}$.The angle $\alpha$ with the positive $x$-axis satisfies $\cos \alpha = \ell = \frac{1}{\sqrt{3}}$.

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

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