If an angle between the line,frac{x + 1}{2} = | vedclass

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(A) The direction ratios of the line are $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} - 2\hat{j} - k

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

Vedclass · Answer

(A) The direction ratios of the line are $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.The normal vector to the plane is $\vec{n} = \hat{i} - 2\hat{j} - k\hat{k}$.The angle $\alpha$ between a line and a plane is given by $\sin \alpha = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.$|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3$.$|\vec{n}| = \sqrt{1^2 + (-2)^2 + (-k)^2} = \sqrt{5 + k^2}$.$\vec{b} \cdot \vec{n} = (2)(1) + (1)(-2) + (-2)(-k) = 2 - 2 + 2k = 2k$.So,$\sin \alpha = \frac{|2k|}{3\sqrt{5 + k^2}}$.Given $\cos \alpha = \frac{2\sqrt{2}}{3}$,we find $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{8}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Equating the two expressions for $\sin \alpha$:$\frac{|2k|}{3\sqrt{5 + k^2}} = \frac{1}{3} \Rightarrow \frac{|2k|}{\sqrt{5 + k^2}} = 1$.Squaring both sides: $\frac{4k^2}{5 + k^2} = 1$.$4k^2 = 5 + k^2 \Rightarrow 3k^2 = 5 \Rightarrow k^2 = \frac{5}{3}$.Thus,$k = \pm \sqrt{\frac{5}{3}}$. Given the options,the correct value is $\sqrt{\frac{5}{3}}$.

If an angle between the line,$\frac{x + 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{-2}$ and the plane,$x - 2y - kz = 3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$,then a value of $k$ is

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