The angle theta between the line x = frac{y-1 | vedclass

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Question

(A) The direction vector of the line is $\vec{b} = (1, 2, \lambda)$ and the normal vector to the plane is $\vec{n} = (1, 2, 3)$.The angle $	heta$ bet

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) The direction vector of the line is $\vec{b} = (1, 2, \lambda)$ and the normal vector to the plane is $\vec{n} = (1, 2, 3)$.The angle $	heta$ between a line and a plane is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Given $	heta = \cos^{-1} \sqrt{\frac{5}{14}}$,we have $\cos 	heta = \sqrt{\frac{5}{14}}$,so $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{5}{14} = \frac{9}{14}$.Thus,$\sin 	heta = \frac{3}{\sqrt{14}}$.Using the formula: $\frac{|(1)(1) + (2)(2) + (3)(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.$\frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.$|5 + 3\lambda| = 3 \sqrt{5 + \lambda^2}$.Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.$25 + 30\lambda + 9\lambda^2 = 45 + 9\lambda^2$.$30\lambda = 20$.$\lambda = \frac{20}{30} = \frac{2}{3}$.

The angle $\theta$ between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 6$ is given by $\cos^{-1} \sqrt{\frac{5}{14}}$. Find the value of $\lambda$.

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