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 - Review Exercises - Page 854: 7

Answer

$$\sqrt {221} $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{u}} = \left\langle {2,4, - 5} \right\rangle {\text{ and }}{\bf{v}} = \left\langle { - 6,10,2} \right\rangle \cr & {\bf{u}} + {\bf{v}} = \left\langle {2,4, - 5} \right\rangle + \left\langle { - 6,10,2} \right\rangle \cr & {\text{Add the vectors}} \cr & {\bf{u}} + {\bf{v}} = \left\langle {2 - 6,4 + 10, - 5 + 2} \right\rangle \cr & {\bf{u}} + {\bf{v}} = \left\langle { - 4,14, - 3} \right\rangle \cr & {\text{Calculate }}\left| {{\bf{u}} + {\bf{v}}} \right| \cr & \left| {{\bf{u}} + {\bf{v}}} \right| = \left| {\left\langle { - 4,14, - 3} \right\rangle } \right| \cr & \left| {{\bf{u}} + {\bf{v}}} \right| = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( {14} \right)}^2} + {{\left( { - 3} \right)}^2}} \cr & \left| {{\bf{u}} + {\bf{v}}} \right| = \sqrt {16 + 196 + 9} \cr & \left| {{\bf{u}} + {\bf{v}}} \right| = \sqrt {221} \cr} $$
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