Calculus: Early Transcendentals 8th Edition

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

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 832: 48

Answer

$(0,7,9)$

Work Step by Step

Vector equation of a line joining the points with position vectors $r_0$ and $r_1$ is: $r=(1-t)r_0+tr_1$ Put $r_0=\lt -3,1,0\gt$ and $r_1= \lt -1,5,6\gt $ $r=(1-t)\lt -3,1,0\gt+t \lt -1,5,6\gt$ $r=\lt -3+2t,1+4t,6t\gt$ Parametric equations of the line are: $x= -3+2t,y=1+4t,z=6t$ As we are given the equation of the plane is: $2x+y-z=-2$ Plug in $x= -3+2t,y=1+4t,z=6t$, we get $t=1.5$ Thus, $x= -3+2(1.5),y=1+4(1.5),z=6(1.5)$ Point of intersection: $(0,7,9)$
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