If the angle between the line x = frac{y - 1} | vedclass

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(C) The line is given by $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$. The direction vector of the line is $\vec{b} = \hat{i} + 2\hat{j

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$\frac{2}{3}$

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(C) The line is given by $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$. The direction vector of the line is $\vec{b} = \hat{i} + 2\hat{j} + \lambda\hat{k}$.The normal vector to the plane $x + 2y + 3z = 4$ is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.Let $	heta$ be the angle between the line and the plane. Then $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Given $	heta = \cos^{-1}\left(\sqrt{\frac{5}{14}}ight)$,we have $\cos 	heta = \sqrt{\frac{5}{14}}$,so $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{5}{14} = \frac{9}{14}$. Thus $\sin 	heta = \frac{3}{\sqrt{14}}$.Now,$\vec{b} \cdot \vec{n} = (1)(1) + (2)(2) + (\lambda)(3) = 1 + 4 + 3\lambda = 5 + 3\lambda$.$|\vec{b}| = \sqrt{1^2 + 2^2 + \lambda^2} = \sqrt{5 + \lambda^2}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.So,$\frac{3}{\sqrt{14}} = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}}$.$3 = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2}} \implies 9(5 + \lambda^2) = (5 + 3\lambda)^2$.$45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2$.$45 = 25 + 30\lambda \implies 30\lambda = 20 \implies \lambda = \frac{20}{30} = \frac{2}{3}$.

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\lambda = \dots$

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