## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 9

#### Answer

$\approx 0.75$ rad

#### Work Step by Step

The angle between two plane can be determined as:: $\theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Here, $a=\lt 2,1,0 \gt$ and $b=\lt 1,2,-1 \gt$ $|a|=\sqrt{2^2+1^2+0^2}= \sqrt {5}$ and $|b|=\sqrt{1^2+2^2+(-1)^2}=\sqrt 6$ Thus, $\theta = \cos ^{-1} (\dfrac{4}{ (\sqrt 5)(\sqrt 6)}) \approx 0.75$ rad

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