A plane Pi passes through the points A=(0,0,2 | vedclass

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(A) The equation of the plane passing through $A(0,0,2)$, $B(1,0,1)$, and $C(3,1,1)$ is given by the determinant equation: $\begin{vmatrix} x-0 & y-0

Vedclass · Accepted Answer

$\frac{7}{6}$

Vedclass · Answer

(A) The equation of the plane passing through $A(0,0,2)$, $B(1,0,1)$, and $C(3,1,1)$ is given by the determinant equation: $\begin{vmatrix} x-0 & y-0 & z-2 \ 1-0 & 0-0 & 1-2 \ 3-0 & 1-0 & 1-2 \end{vmatrix} = 0$ $\Rightarrow \begin{vmatrix} x & y & z-2 \ 1 & 0 & -1 \ 3 & 1 & -1 \end{vmatrix} = 0$ Expanding along the first row: $x(0 - (-1)) - y(-1 - (-3)) + (z-2)(1 - 0) = 0$ $x(1) - y(2) + (z-2)(1) = 0$ $x - 2y + z - 2 = 0$. The normal vector to the plane is $\vec{n} = \langle 1, -2, 1 angle$. The $XY$-plane has normal $\vec{n}_1 = \langle 0, 0, 1 angle$. The angle $\alpha$ between the planes is given by $\cos \alpha = \frac{|\vec{n} \cdot \vec{n}_1|}{|\vec{n}| |\vec{n}_1|} = \frac{|1|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{1^2}} = \frac{1}{\sqrt{6}}$. Thus, $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{6} = \frac{5}{6}$. The $XZ$-plane has normal $\vec{n}_2 = \langle 0, 1, 0 angle$. The angle $\beta$ between the planes is given by $\cos \beta = \frac{|\vec{n} \cdot \vec{n}_2|}{|\vec{n}| |\vec{n}_2|} = \frac{|-2|}{\sqrt{6} \sqrt{1^2}} = \frac{2}{\sqrt{6}}$. Thus, $\sin^2 \beta = 1 - \cos^2 \beta = 1 - \frac{4}{6} = \frac{2}{6}$. Therefore, $\sin^2 \alpha + \sin^2 \beta = \frac{5}{6} + \frac{2}{6} = \frac{7}{6}$.

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

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