If a line L is common to the planes x-y+z+2=0 | vedclass

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(A) The line $L$ is the intersection of the two planes $x-y+z+2=0$ and $2x+y-2z+5=0$.The normal vectors to these planes are $\vec{n_1} = \hat{i} - \ha

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$\left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}}\right)$

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(A) The line $L$ is the intersection of the two planes $x-y+z+2=0$ and $2x+y-2z+5=0$.The normal vectors to these planes are $\vec{n_1} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction vector $\vec{v}$ of the line $L$ is perpendicular to both normals, so $\vec{v} = \vec{n_1} 	imes \vec{n_2}$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(-2-2) + \hat{k}(1+2) = \hat{i} + 4\hat{j} + 3\hat{k}$.The direction ratios of the line are $(1, 4, 3)$.The magnitude of the direction vector is $\sqrt{1^2 + 4^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.The direction cosines are given by $\left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}}ight)$.

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$, then the direction cosines of the line $L$ are

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