The plane passing through the line L: ell x-y | vedclass

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(C) The normal vectors to the planes forming the line $L$ are $\vec{n}_{1} = \ell \hat{i} - \hat{j} + 3(1-\ell) \hat{k}$ and $\vec{n}_{2} = \hat{i} +

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$125$

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(C) The normal vectors to the planes forming the line $L$ are $\vec{n}_{1} = \ell \hat{i} - \hat{j} + 3(1-\ell) \hat{k}$ and $\vec{n}_{2} = \hat{i} + 2\hat{j} - \hat{k}$.The direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n}_{1} 	imes \vec{n}_{2}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \ell & -1 & 3(1-\ell) \ 1 & 2 & -1 \end{vmatrix} = (1 - 6 + 6\ell) \hat{i} - (-\ell - 3 + 3\ell) \hat{j} + (2\ell + 1) \hat{k} = (6\ell - 5) \hat{i} + (3 - 2\ell) \hat{j} + (2\ell + 1) \hat{k}$.The plane $3x - 8y + 7z = 4$ contains the line $L$,so the normal vector of this plane,$\vec{n}_{3} = 3\hat{i} - 8\hat{j} + 7\hat{k}$,must be perpendicular to the direction vector $\vec{v}$ of the line.Thus,$\vec{v} \cdot \vec{n}_{3} = 0$:$3(6\ell - 5) - 8(3 - 2\ell) + 7(2\ell + 1) = 0$$18\ell - 15 - 24 + 16\ell + 14\ell + 7 = 0$$48\ell - 32 = 0 \implies \ell = \frac{32}{48} = \frac{2}{3}$.Substituting $\ell = \frac{2}{3}$ into $\vec{v}$:$\vec{v} = (6(\frac{2}{3}) - 5) \hat{i} + (3 - 2(\frac{2}{3})) \hat{j} + (2(\frac{2}{3}) + 1) \hat{k} = -1\hat{i} + \frac{5}{3}\hat{j} + \frac{7}{3}\hat{k}$.The angle $	heta$ between the line $L$ and the $y$-axis (direction $\hat{j}$) is given by $\cos 	heta = \frac{|\vec{v} \cdot \hat{j}|}{|\vec{v}| |\hat{j}|}$.$|\vec{v}| = \sqrt{(-1)^2 + (\frac{5}{3})^2 + (\frac{7}{3})^2} = \sqrt{1 + \frac{25}{9} + \frac{49}{9}} = \sqrt{\frac{9+25+49}{9}} = \sqrt{\frac{83}{9}} = \frac{\sqrt{83}}{3}$.$\cos 	heta = \frac{5/3}{\sqrt{83}/3} = \frac{5}{\sqrt{83}}$.Therefore,$415 \cos^{2} 	heta = 415 	imes \frac{25}{83} = 5 	imes 25 = 125$.

The plane passing through the line $L: \ell x-y+3(1-\ell)z=1, x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis,then $415 \cos^{2} \theta$ is equal to...

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