Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - Review Exercises - Page 811: 7

Answer

$$d = \sqrt {22} $$

Work Step by Step

$$\eqalign{ & {\text{Let the points be }}\underbrace {\left( {1,6,3} \right)}_{\left( {{x_1},{y_1},{z_1}} \right)},\underbrace {\left( { - 2,3,5} \right)}_{\left( {{x_2},{y_2},{z_2}} \right)} \cr & {\text{Finding the Distance Between Two Points}} \cr & d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \cr & d = \sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( {3 - 6} \right)}^2} + {{\left( {5 - 3} \right)}^2}} \cr & {\text{Simplifying}} \cr & d = \sqrt {9 + 9 + 4} \cr & d = \sqrt {22} \cr} $$
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