Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 5 - Inner Product Spaces - 5.1 Definition of an Inner Product Space - Problems - Page 350: 2


$$ \theta =\cos^{-1}\frac{8}{3 \sqrt{21} }$$

Work Step by Step

Given $$v=(1,3,-1,4),w=(-1,1,-2,1)$$ Let $\theta$ be an angle between the vectors $v$ and $w .$ We know that $$\cos \theta=\frac{\langle v,w\rangle}{\|v\|\|w\|}$$ . Since, we have \begin{aligned} \langle v, w\rangle&= 1\cdot(-1)+(3) \cdot 1+(-1) \cdot(-2)+4 \cdot (1)\\ &=8\\ \|v\|&=\sqrt{\langle v, v\rangle}\\ &=\sqrt{1 \cdot 1+3 \cdot 3+(-1) \cdot(-1)+4 \cdot 4}\\ &=\sqrt{1+9+1+16}\\ &=\sqrt{27}\\ &=3 \sqrt{3}\\ \|w\|&=\sqrt{\langle w,w\rangle}\\ &=\sqrt{(-1) \cdot(-1)+1 \cdot 1+(-2) \cdot(-2)+1 \cdot 1}\\ &=\sqrt{1+1+4+1}\\ &=\sqrt{7} \end{aligned} and So we get: $$\cos \theta=\frac{\langle v,w\rangle}{\|v\|\|w\|}=\frac{8}{3 \sqrt{3} \sqrt{7}}=\frac{8}{3 \sqrt{21} }$$ The angle between $v$ and $w$ is $\cos^{-1}\frac{8}{3 \sqrt{21} }$
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