Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 51

$0.82$

The angle between two plane is given by: $ \theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Here, $a=\lt 2,2,-1 \gt$ and $b=\lt 1,2,1 \gt$ $|a|=\sqrt{2^2+2^2+(-1)^2}= 3$ and $|b|=\sqrt{1^2+2^2+1^2}=\sqrt{1+4+1}=\sqrt 6$ Now, $ \theta = \cos ^{-1} (\dfrac{p \cdot q}{|p||q|})=\cos ^{-1} (\dfrac{5}{ 3 \sqrt 6})=0.82$