Let P(1, -2, 5) be the foot of the perpendicu | vedclass

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(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector from the origin $(0, 0, 0)$ to $P(1, -2, 5)$, which is $\vec{n_1} = (1, -2, 5)$.T

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$\cos^{-1}\left(\frac{19}{\sqrt{390}}\right)$

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(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector from the origin $(0, 0, 0)$ to $P(1, -2, 5)$, which is $\vec{n_1} = (1, -2, 5)$.The normal vector $\vec{n_2}$ to the plane $\pi_2$ is the vector from $(1, 2, -1)$ to $P(1, -2, 5)$, which is $\vec{n_2} = (1-1, -2-2, 5-(-1)) = (0, -4, 6)$.The acute angle $	heta$ between the two planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(0) + (-2)(-4) + (5)(6) = 0 + 8 + 30 = 38$.Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.$|\vec{n_2}| = \sqrt{0^2 + (-4)^2 + 6^2} = \sqrt{0 + 16 + 36} = \sqrt{52} = 2\sqrt{13}$.Thus, $\cos 	heta = \frac{38}{\sqrt{30} \cdot 2\sqrt{13}} = \frac{19}{\sqrt{390}}$.Therefore, $	heta = \cos^{-1}\left(\frac{19}{\sqrt{390}}ight)$.

Let $P(1, -2, 5)$ be the foot of the perpendicular drawn from the origin to the plane $\pi_1$ and the same $P$ be the foot of the perpendicular from $(1, 2, -1)$ to the plane $\pi_2$. Then the acute angle between the planes $\pi_1$ and $\pi_2$ is

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