Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.3 Exercises - Page 831: 59

Answer

Property $2$: $\vec {a} \cdot \vec {b}=\vec {b} \cdot \vec {a}$ Property $4$: $(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$ Property $5$: $\vec {0} \cdot \vec {a}=0$

Work Step by Step

Proof of Property $2$. Let $a= \lt a_1,a_2,a_3 \gt$ and $b= \lt b_1,b_2,b_3 \gt$ $\vec {a} \cdot \vec {b}=a_1b_1+a_2b_2+a_3b_3$ Using commutative law: $\vec {a} \cdot \vec {b}=a_1b_1+a_2b_2+a_3b_3=b_1a_1+b_2a_2+b_3a_3$ Thus, $\vec {a} \cdot \vec {b}=\vec {b} \cdot \vec {a}$ Proof of Property $4$. Let $a= \lt a_1,a_2,a_3 \gt$ and $b= \lt b_1,b_2,b_3 \gt$ $(c\vec {a}) \cdot \vec {b}=ca_1b_1+ca_2b_2+ca_3b_3=c(a_1b_1+a_2b_2+a_3b_3) $ $(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$ Thus,$(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$ Proof of Property $5$. Let $0= \lt 0,0,0 \gt$ and $a= \lt a_1,a_2,a_3 \gt$ $\vec {0} \cdot \vec {a}=\lt 0,0,0 \gt \cdot \lt a_1,a_2,a_3 \gt$ $\vec {0} \cdot \vec {a}=0a_1+0a_2+0a_3$ Thus, $\vec {0} \cdot \vec {a}=0$
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