Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.2 - Vectors - Exercises 12.2 - Page 705: 41


$a=\dfrac{3}{2}$ and $b=\dfrac{1}{2}$

Work Step by Step

Let us consider that $u=2i+j,v=i+j,w=i-j$ Since, $u=av+bw$ and $2i+j=ai+aj+bi-bj$ Equate the terms. $a+b=2; a-b=1$ Solve the above two equations, we get $2a=3$ or, $a=\dfrac{3}{2}$ Now, $a+b=2$ or, $\dfrac{3}{2}+b=2$ and $b=\dfrac{1}{2}$ Hence, we have $a=\dfrac{3}{2}$ and $b=\dfrac{1}{2}$
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