Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 726: 10


$x=2-2t,y=3+4t, z=-2t$

Work Step by Step

As we know the parametric equations of a straight line for a vector $v=v_1i+v_2j+v_3k$ passing through a point $P(a,b,c)$ is given by $x=a+t v_1,y=b+t v_2; z=c+t v_3$ Consider two vectors u and v and line is perpendicular to these vectors.This implies that these vectors are parallel to the cross product of $n=u \times v$ and $n=-2i+4j-2k$ So, we have $v=\lt -2,4,-2 \gt$ and $P=(2,3,0)$ . Our parametric equations are: $x=2-2t,y=3+4t, z=-2t$
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