Let A=(2,0,-1),B=(1,-2,0),C=(1,2,-1),and D=(0 | vedclass

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(C) First,find the normal vector $\vec{n_1}$ to the plane $ABC$. The vectors $\vec{AB} = B-A = (-1, -2, 1)$ and $\vec{AC} = C-A = (-1, 2, 0)$.$\vec{n_

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$\frac{3}{\sqrt{5}}$

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(C) First,find the normal vector $\vec{n_1}$ to the plane $ABC$. The vectors $\vec{AB} = B-A = (-1, -2, 1)$ and $\vec{AC} = C-A = (-1, 2, 0)$.$\vec{n_1} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -2 & 1 \ -1 & 2 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0 - (-1)) + \hat{k}(-2 - 2) = -2\hat{i} - \hat{j} - 4\hat{k}$.Next,find the normal vector $\vec{n_2}$ to the plane $ACD$. The vectors $\vec{AC} = (-1, 2, 0)$ and $\vec{AD} = D-A = (-2, -1, -1)$.$\vec{n_2} = \vec{AC} 	imes \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 0 \ -2 & -1 & -1 \end{vmatrix} = \hat{i}(-2-0) - \hat{j}(1-0) + \hat{k}(1 - (-4)) = -2\hat{i} - \hat{j} + 5\hat{k}$.The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\vec{n_1} \cdot \vec{n_2} = (-2)(-2) + (-1)(-1) + (-4)(5) = 4 + 1 - 20 = -15$.$|\vec{n_1}| = \sqrt{(-2)^2 + (-1)^2 + (-4)^2} = \sqrt{4+1+16} = \sqrt{21}$.$|\vec{n_2}| = \sqrt{(-2)^2 + (-1)^2 + 5^2} = \sqrt{4+1+25} = \sqrt{30}$.$\cos 	heta = \frac{|-15|}{\sqrt{21} \sqrt{30}} = \frac{15}{\sqrt{630}} = \frac{15}{3\sqrt{70}} = \frac{5}{\sqrt{70}}$.Since $\cos^2 	heta = \frac{25}{70} = \frac{5}{14}$,then $\sin^2 	heta = 1 - \frac{5}{14} = \frac{9}{14}$.Thus,$	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta} = \frac{9/14}{5/14} = \frac{9}{5}$.Therefore,$	an 	heta = \frac{3}{\sqrt{5}}$.

Let $A=(2,0,-1)$,$B=(1,-2,0)$,$C=(1,2,-1)$,and $D=(0,-1,-2)$ be four points. If $\theta$ is the acute angle between the plane determined by $A, B, C$ and the plane determined by $A, C, D$,then $\tan \theta=$

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