Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 50

$0.96$ 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