A line L passes through the points (1, 2, -3) | vedclass

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(D) The direction vector of line $L$ passing through $(1, 2, -3)$ and $(3, 3, -1)$ is $\vec{v} = (3-1, 3-2, -1-(-3)) = (2, 1, 2)$.The equation of the

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(D) The direction vector of line $L$ passing through $(1, 2, -3)$ and $(3, 3, -1)$ is $\vec{v} = (3-1, 3-2, -1-(-3)) = (2, 1, 2)$.The equation of the plane $\pi$ passing through $(2, 1, -2), (-2, -3, 6)$,and $(0, 2, -1)$ is given by the determinant:$\begin{vmatrix} x-2 & y-1 & z+2 \ -2-2 & -3-1 & 6+2 \ 0-2 & 2-1 & -1+2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-2 & y-1 & z+2 \ -4 & -4 & 8 \ -2 & 1 & 1 \end{vmatrix} = 0$.Expanding the determinant: $(x-2)(-4-8) - (y-1)(-4+16) + (z+2)(-4-8) = 0$.$-12(x-2) - 12(y-1) - 12(z+2) = 0 \Rightarrow x-2 + y-1 + z+2 = 0 \Rightarrow x+y+z = 1$.The normal vector to the plane is $\vec{n} = (1, 1, 1)$.The angle $	heta$ between the line and the plane is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.$\sin 	heta = \frac{|(2)(1) + (1)(1) + (2)(1)|}{\sqrt{2^2+1^2+2^2} \sqrt{1^2+1^2+1^2}} = \frac{|2+1+2|}{\sqrt{9} \sqrt{3}} = \frac{5}{3\sqrt{3}}$.Therefore,$\sin^2 	heta = \frac{25}{9 	imes 3} = \frac{25}{27}$.Since $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{25}{27} = \frac{2}{27}$.Thus,$27 \cos^2 	heta = 2$.

$A$ line $L$ passes through the points $(1, 2, -3)$ and $(3, 3, -1)$ and a plane $\pi$ passes through the points $(2, 1, -2), (-2, -3, 6)$,and $(0, 2, -1)$. If $\theta$ is the angle between the line $L$ and plane $\pi$,then $27 \cos^2 \theta = $

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