## Thomas' Calculus 13th Edition

$\dfrac{\pi}{2}$ or $90 ^{\circ}$
we have, $\overrightarrow{AC}=\lt 2,4 \gt$ and $\overrightarrow{DB}=\lt -4,2 \gt$ The angle between two planes can be determined as: $\theta = \cos ^{-1} (\dfrac{\overrightarrow{AC} \cdot \overrightarrow{DB}}{|\overrightarrow{AC}||\overrightarrow{DB}|})$ Now, $|\overrightarrow{AC}|=\sqrt{(2)^2+(4)^2}= \sqrt {20}$ and $|\overrightarrow{DB}|=\sqrt{(-4)^2+(2)^2}=\sqrt {20}$ Thus, $\theta =\cos ^{-1} (\dfrac{(2)(-4)+(4)(2)}{ ( \sqrt {20})(\sqrt {20})})$ or, $\cos ^{-1} (0)=\dfrac{\pi}{2}$ or $90 ^{\circ}$