Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 853: 26

Answer

$x=1 \pm \frac{ \sqrt {6}}{2}$

Work Step by Step

As we know $a \cdot b = |a| |b| cos \theta$ Here, $a \cdot b= \lt 2,1,-1 \gt \cdot \lt 1,x,0 \gt=2+x$ $2+x=(\sqrt 6) (\sqrt {1+x^2})cos 45^\circ$ $2+x=(\sqrt 6) (\sqrt {1+x^2})\dfrac{\sqrt 2}{2}$ $(2+x)^2=(\sqrt 3\sqrt {1+x^2})^2$ $x^2+4+2x=(\sqrt 3\sqrt {1+x^2})^2$ $x^2+4+2x=3+3x^2$ $2x^2-4x-1=0$ $x=\frac{4 \pm \sqrt {24}}{4}$ Hence, $x=1 \pm \frac{ \sqrt {6}}{2}$
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