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

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(D) The line is given by $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$. The direction vector of the line is $\vec{b} = (1, 2, \lambda)$.The no

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$\frac{2}{3}$

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(D) The line is given by $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$. The direction vector of the line is $\vec{b} = (1, 2, \lambda)$.The normal vector to the plane $x + 2y + 3z = 4$ is $\vec{n} = (1, 2, 3)$.Let $	heta$ be the angle between the line and the plane. The formula for the angle is $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Given $	heta = \cos^{-1}\left(\sqrt{\frac{5}{14}}ight)$,we have $\cos 	heta = \sqrt{\frac{5}{14}}$.Using $\sin^2 	heta + \cos^2 	heta = 1$,we get $\sin 	heta = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}}$.Now,$\frac{|(1)(1) + (2)(2) + (3)(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.$\frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.$|5 + 3\lambda| = 3\sqrt{5 + \lambda^2}$.Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.$25 + 9\lambda^2 + 30\lambda = 45 + 9\lambda^2$.$30\lambda = 20$.$\lambda = \frac{20}{30} = \frac{2}{3}$.

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 $\lambda$ equals:

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