Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 666: 3

Answer

41

Work Step by Step

We know that the scalar product of $a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$ and $b_{1}\hat{i}+b_{2}\hat{j}+ b_{3}\hat{k}$ is equal to $a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$. Therefore, $<0,1,0>\cdot<7,41,-3>=$$0\times7+1\times4+0\times(-3)= 41$
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