Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Chapter Test - Page 636: 16

Answer

$\frac{\sqrt2i}{2}-\frac{\sqrt2j}{2}$.

Work Step by Step

If a vector $v$ initiates at point $P_1(x_1,y_1)$ and terminates at $P_2(x_2,y_2)$ then $v$ is equal to the position vector $v=(x_2-x_1)i+(y_2-y_1)j.$ Hence here: $v=(8\sqrt2-3\sqrt2)i+(2\sqrt2-7\sqrt2)j=5\sqrt2i-5\sqrt2j.$ The magnitude $||v||$ of a vector $v=ai+bj$, can be computed using the formula $||v||=\sqrt{a^2+b^2}$. Hence here: $||v||=\sqrt{(5\sqrt2)^2+(-5\sqrt2)^2}=\sqrt{50+50}=\sqrt{100}=10$. If $v=ai+bj$ then the unit vector $u$ in the direction of $v$ is $u=\frac{v}{||v||}$. Hence here: $u=\frac{5\sqrt2i-5\sqrt2j}{10}=\frac{\sqrt2i}{2}-\frac{\sqrt2j}{2}$.
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