# Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 10

$\approx 0.84$ rad

#### Work Step by Step

The angle between two plane can be determined as:: $\theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Now $a=\lt 2,-2,1 \gt$ and $b=\lt 3,0,4 \gt$ Here, $|a|=\sqrt{2^2+(-2)^2+1^2}= \sqrt {9}=3$ and $|b|=\sqrt{3^2+0^2+(4)^2}=\sqrt {25}=5$; so, $\theta = \cos ^{-1} (\dfrac{10}{ (3)(5)})=\cos ^{-1} (\dfrac{10}{15} )\approx 0.84$ rad

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