## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.3 - The Dot Product - Exercises - Page 616: 11

#### Answer

$1.77$ rad

#### Work Step by Step

The formula to calculate the angle between two planes is: $\theta = \cos ^{-1} (\dfrac{p \cdot q}{|p||q|})$ Here, $p=\lt \sqrt 3,-7,0 \gt$ and $q=\lt \sqrt 3,1,-2 \gt$ $|p|=\sqrt{(\sqrt 3)^2+(-7)^2+(0)^2}= 2 \sqrt {13}$ and $|q|=\sqrt{(\sqrt 3)^2+(1)^2+(-2)^2}=\sqrt {8}$ Thus, $\theta = \cos ^{-1} (\dfrac{p \cdot q}{|p||q|})=\cos ^{-1} \dfrac{-4}{ ( 2 \sqrt {13})(\sqrt 8)})=\cos ^{-1} \dfrac{-4}{ 4 \sqrt {26}}$ or, $\theta \approx 1.77$ rad

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