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: 78

Answer

$2$

Work Step by Step

For line 1: $x=1+t,y=1+6t,z=2t$ For line 2: $x=1+2s,y=5+15s,z=-2+6s$ Direction vector of first line is$v_1=\lt 1,6,2\gt $ and Direction vector of second line is$v_2=\lt 2,15,6\gt $ Two skew lines are lying on two parallel planes. $n=v_1 \times v_2=\lt 6,-2,3\gt $ Letting $s=0$ for line 2:$x=1+2s,y=5+15s,z=-2+6s$, we get the points $(1,5,-2)$ From the plane equation, we have $6x-2y+3z+10=0$ Now, letting $t=1$ for line 1, we get the points $(1,1,0)$ Distance formula between a point and a plane is given as: $D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|6(1)-2(1)+3(0)+10|}{\sqrt {6^2+(-2)^2+3^2}}$ $=\frac{14}{7}$ $=2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.