Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 11

$\approx 1.77$ rad

The angle between two plane can be determined as: $ \theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Now, $a=\lt \sqrt 3,-7,0 \gt$ and $b=\lt \sqrt 3,1,-2 \gt$ Here, $|a|=\sqrt{(\sqrt 3)^2+(-7)^2+(0)^2}= 2 \sqrt {13}$ and $|b|=\sqrt{(\sqrt 3)^2+(1)^2+(-2)^2}=\sqrt {8}$ So, $ \theta = \cos ^{-1} (\dfrac{-4}{ ( 2 \sqrt {13})(\sqrt 8)})\approx 1.77$ rad