## Thomas' Calculus 13th Edition

$\dfrac{\pi}{2}$
The angle between two plane is given as $\theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Here, $a=\lt 5,1,-1\gt$ and $b=\lt 1,-2,3 \gt$ $|a|=\sqrt{5^2+1^2+(-1)^2}=3 \sqrt 3$ and $|b|=\sqrt{1^2+(-2)^2+3^2}=\sqrt {14}$ Thus, $\theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})=\cos ^{-1} (\dfrac{0}{(3 \sqrt 3) (\sqrt {14})})$ or, $=\cos ^{-1} (0)$ Thus, $\theta =\dfrac{\pi}{2}$