## Thomas' Calculus 13th Edition

$0.96$ rad
The angle between two plane is given by: $\theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Here, $a=\lt 1,1,1 \gt$ and $b=\lt 0,0,1 \gt$ $|a|=\sqrt{1^2+1^2+1^2}= \sqrt 3$ and $|b|=\sqrt{0^2+0^2+1^2}=\sqrt{1}=1$ Hence, $\theta =cos ^{-1} (\dfrac{1}{ \sqrt 3})=0.96$ rad