Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 873: 79


$\frac{13}{\sqrt {69}}$

Work Step by Step

Line 1 goes from the origin and $(2,0,-1)$ Line 2 goes from the $(1,-1,1)$ and $(4,1,3)$ Direction vector of first line is$v_1=\lt 2,0,-1\gt $ and Direction vector of second line is$v_2=\lt 3,2,2 \gt $ Two skew lines are lying on two parallel planes. $n=v_1 \times v_2=\lt 2,-7,4\gt $ Use the points $(1,-1,1)$ from line 2 and $n$ to form the plane equation. From the plane equation, we have $2x-7y+4z-13=0$ Use points $(0,0,0)$ for line 1. Distance formula between a point and a plane is given as: $D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|2(0)-7(0)+4(0)-13|}{\sqrt {2^2+(-7)^2+4^2}}$ $=\frac{13}{\sqrt {69}}$
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