The angle between the line frac{x+1}{2}=frac{ | vedclass

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(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left| \frac

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$\sqrt{\frac{5}{3}}$

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(C) The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} ight|$.Here,the direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$ and the normal vector of the plane is $\vec{n} = \hat{i} - 2\hat{j} - \lambda\hat{k}$.Given $	heta = \cos^{-1}\left(\frac{2\sqrt{2}}{3}ight)$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - \left(\frac{2\sqrt{2}}{3}ight)^2} = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Substituting the values into the formula:$\frac{1}{3} = \left| \frac{(2)(1) + (1)(-2) + (-2)(-\lambda)}{\sqrt{2^2 + 1^2 + (-2)^2} \sqrt{1^2 + (-2)^2 + (-\lambda)^2}} ight|$$\frac{1}{3} = \left| \frac{2 - 2 + 2\lambda}{\sqrt{4 + 1 + 4} \sqrt{1 + 4 + \lambda^2}} ight|$$\frac{1}{3} = \left| \frac{2\lambda}{3 \sqrt{5 + \lambda^2}} ight|$Squaring both sides:$\frac{1}{9} = \frac{4\lambda^2}{9(5 + \lambda^2)}$$5 + \lambda^2 = 4\lambda^2$$3\lambda^2 = 5$$\lambda^2 = \frac{5}{3} \Rightarrow \lambda = \sqrt{\frac{5}{3}}$.

The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane $x-2y-\lambda z=3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then the value of $\lambda$ is:

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