Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.1 Vectors in the Plane - 11.1 Exercises - Page 769: 69

Answer

$(a,b)=\dfrac{a+b}{2}u+\dfrac{b-a}{2}v$

Work Step by Step

We have: $(a,b)=c_1u+c_2v$ $(a,b)=c_1(1,1)+c_2(-1,1)$ $(a,b)=(c_1,c_1)+(-c_2,c_2)$ $(a,b)=(c_1-c_2,c_1+c_2)$ Determine $c_1$ and $c_2$: $\begin{cases} c_1-c_2=a\\ c_1+c_2=b \end{cases}$ $c_1-c_2+c_1+c_2=a+b$ $2c_1=a+b$ $c_1=\dfrac{a+b}{2}$ $c_2=b-c_1=b-\dfrac{a+b}{2}=\dfrac{2b-a-b}{2}=\dfrac{b-a}{2}$ The vector can be written: $(a,b)=\dfrac{a+b}{2}u+\dfrac{b-a}{2}v$
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