The acute angle theta between the xy-plane an | vedclass

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(A) Step $1$: Find the normal vector $\vec{n_1}$ to the plane. The plane is parallel to $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{

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$\cos^{-1} \left( \frac{8}{5\sqrt{5}} \right)$

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(A) Step $1$: Find the normal vector $\vec{n_1}$ to the plane. The plane is parallel to $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$.Step $2$: $\vec{n_1} = \vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & -1 \ 1 & -2 & -2 \end{vmatrix} = \hat{i}(-4 - 2) - \hat{j}(-6 + 1) + \hat{k}(-6 - 2) = -6\hat{i} + 5\hat{j} - 8\hat{k}$.Step $3$: The normal to the $xy$-plane is $\vec{n_2} = \hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.Step $4$: The angle $	heta$ between the planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.Step $5$: $\vec{n_1} \cdot \vec{n_2} = (-6)(0) + (5)(0) + (-8)(1) = -8$. So, $|\vec{n_1} \cdot \vec{n_2}| = 8$.Step $6$: $|\vec{n_1}| = \sqrt{(-6)^2 + 5^2 + (-8)^2} = \sqrt{36 + 25 + 64} = \sqrt{125} = 5\sqrt{5}$. $|\vec{n_2}| = 1$.Step $7$: $\cos 	heta = \frac{8}{5\sqrt{5}}$. Thus, $	heta = \cos^{-1} \left( \frac{8}{5\sqrt{5}} ight)$.

The acute angle $\theta$ between the $xy$-plane and the plane passing through the point $(1, 2, 4)$ and parallel to the vectors with direction ratios $3, 2, -1$ and $1, -2, -2$ is

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