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 - 11.1 Exercises: 73

Answer

$$\mathbf{v}=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$$

Work Step by Step

$(\mathbf{u}+\mathbf{v})_x=u_x+v_x$, and $(\mathbf{u}+\mathbf{v})_y=u_y+v_y$ $u_x=\|\mathbf{u}\|\cos{\theta_u}=1\cos{45^\circ}=\frac{\sqrt{2}}{2}$ $u_y=\|\mathbf{u}\|\sin{\theta_u}=1\sin{45^\circ}=\frac{\sqrt{2}}{2}$ $(\mathbf{u}+\mathbf{v})_x=\|(\mathbf{u}+\mathbf{v})\|\cos{\theta_{u+v}}=\sqrt{2}\cos{90^\circ}=0$ $(\mathbf{u}+\mathbf{v})_y=\|(\mathbf{u}+\mathbf{v})\|\sin{\theta_{u+v}}=\sqrt{2}\sin{90^\circ}=\sqrt{2}$ $v_x=(\mathbf{u}+\mathbf{v})_x-u_x=0-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}$ $v_y=(\mathbf{u}+\mathbf{v})_y-u_y=\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$
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