University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.3 - The Dot Product - Exercises - Page 617: 20

Answer

See below.

Work Step by Step

We have two side vectors $u$ and $v$ with two diagonals $d_1=u+v$ and $d_2=u-v$ We know that for any rectangle if $d_1 \perp d_2$, then $d_1 \cdot d_2=0$ so, we get $(-v+(-u)) \cdot (-v+u)=v \cdot v-v \cdot u+u \cdot v -u \cdot u=0$ Now, $v \cdot v-v \cdot u+u \cdot v -u \cdot u=0$ or, $|v|^2-|u|^2=0$ . But $|v| + |u| \ne 0$ Thus, $|v|-|u|=0 \implies |v|=|u|$ Hence, we have a rectangle with two equal adjacent sides -- a square.
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