Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 5 - Inner Product Spaces - 5.5 Applications of Inner Product Spaces - 5.5 Exercises - Page 283: 51

Answer

$4.5\sqrt6$

Work Step by Step

$u=(1,3,5)-(3,3,0)=(-2,0,5)$ $v=(-2,0,5)-(3,3,0)=(-5,-3,5)$ are two sides of the triangle. We know that for a matrix $ \left[\begin{array}{rrr} a & b & c \\ d &e & f \\ g &h & i \\ \end{array} \right] $ the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$ $u×v$ is the determinant of the matrix $\begin{bmatrix} i& j & k \\ -2& 0&5\\ -5&-3 &5 \\ \end{bmatrix} $ Thus $u×v=(15,-15,6).$ Thus $A=0.5|u\times v|=0.5\sqrt{15^2+(-15)^2+6^2}=0.5\sqrt{486}=4.5\sqrt6$
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