University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.3 - The Dot Product - Exercises - Page 617: 28

Answer

No. Counterexample given below.

Work Step by Step

Counterexample: ${\bf u}=\langle 1,1\rangle$ ${\bf v_{1}}=\langle 2,0\rangle,\quad {\bf v_{2}}=\langle 0,2\rangle$ ${\bf u}\cdot{\bf v_{1}}=2+0=2,$ ${\bf u}\cdot{\bf v_{2}}=0+2=2.$ ${\bf u}\cdot{\bf v_{1}}={\bf u}\cdot{\bf v_{2}}$, but ${\bf v_{1}}\neq{\bf v_{2}}$
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