Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Review - Exercises - Page 842: 8

Answer

$(a \times b) .[(b \times c) \times (c \times a)]=[a \cdot (b\times c)] ^2$

Work Step by Step

Use property $a \times (b \times c)=(a.c)b-(a.b)c$ $(a \times b) .[(b \times c) \times (c \times a)]=(a \times b) .[((b \times c)\cdot a)c-((b \times c)c)a]$ Since, the vectors $(b \times c)$ and $c$ are perpendicular, thus, $((b \times c)c)a]=0$ $(a \times b) .[(b \times c) \times (c \times a)]=(a \times b) .[((b \times c)\cdot a)c]$ $(a \times b) .[(b \times c) \times (c \times a)]=(a \times b) .[(a(b \times c)) c]$ $(a \times b) .[(b \times c) \times (c \times a)]=(a (b\times c)) \cdot (a (b\times c)) $ Hence, $(a \times b) .[(b \times c) \times (c \times a)]=[a \cdot (b\times c)] ^2$
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