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 852: 12


$u.v=\frac{1}{2}$ and $u.w=0$

Work Step by Step

The given shape is an equilateral triangle, thus, $|u|=1$ and $|v| $ is a leg of a $45^\circ-45^\circ-90^\circ$ triangle where the hypotenuse is 1, we know it is $|u|=\frac{1}{\sqrt 2}$ Here, $\theta=45^\circ$ As we know $u.v=|u||v|cos\theta$ $u.v=|1||\frac{1}{\sqrt 2}|cos45^\circ$ $=1\times\frac{1}{\sqrt 2}\times\frac{1}{\sqrt 2}$ $=\frac{1}{2}$ We now that the dot product is always $0$ when two vectors are orthogonal and $u$ and $w$ are orthogonal in the given shape. Therefore, $u.w=0$ Hence, $u.v=\frac{1}{2}$ and $u.w=0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.