## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Vectors - Exercises - Page 608: 16

#### Answer

$\lt \dfrac{-1}{ \sqrt 2},\dfrac{1}{\sqrt 2} \gt$

#### Work Step by Step

The coordinates of a point P on the circle with radius $r$ are: $(r \cos \theta, r \sin \theta)$ Therefore, $\lt 1*\cos (135^{\circ}), 1*\sin (135^{\circ}) \gt=\lt \dfrac{-1}{ \sqrt 2},\dfrac{1}{\sqrt 2} \gt$

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