Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.5 Exercises - Page 850: 71

Answer

$\frac{18}{7}$

Work Step by Step

Given: $(1,-2,4)$ ; $3x+2y+6z=5$ Distance formula between the point and the plane is given as: $D=\frac{|ax+by+cz+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|3.1+2 \cdot (-2)+6 \cdot (4)-5|}{\sqrt {3^2+2^2+6^2}}$ $=\frac{18}{7}$
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