Find the angle between the planes ax + by + d | vedclass

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(D) The equation of the first plane is $ax + by + 0z + d = 0$. The normal vector to this plane is $\vec{n_1} = (a, b, 0)$.The equation of the second p

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$\frac{\pi}{2}$

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(D) The equation of the first plane is $ax + by + 0z + d = 0$. The normal vector to this plane is $\vec{n_1} = (a, b, 0)$.The equation of the second plane is $0x + 0y + 1z = 0$. The normal vector to this plane is $\vec{n_2} = (0, 0, 1)$.The angle $	heta$ between 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: $\vec{n_1} \cdot \vec{n_2} = (a)(0) + (b)(0) + (0)(1) = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

Find the angle between the planes $ax + by + d = 0$ $(a^2 + b^2 \neq 0)$ and $z = 0$.

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