Answer to Question #109924 in Analytic Geometry for Joseph Se

Question

Answer the two questions and show the complete solutions.

4.Show that the line x/3 = y/-2 = z/2 is parallel to the plane 2

Answer 1

firstly , the plane equation is missing ,so , i will prove this statement on the following plane equation

$2x+2y-z=6$

Since , If a line is parallel to a plane, it will be perpendicular to the plane’s normal vector , the equation of a plane having the following normal vector 

$<2,2,-1>$

On the other hand from line equation we can write it in the parametric form

$x=3t,\ \ \ \ y=-2t ,\ \ \ \ z= 2t$

Hence  the line’s vector is

$<3,-2,2>$

Since the dot product between the normal vector and the line vector given by

$<2,2,-1> \cdot <3,-2,2>= (2)(3)+(2)(-2)+(-1)(2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6-4-2=0$

This means that the normal vector and the line vector are perpendicular , hence the line

$\frac{x}{3}=\frac{y}{-2} = \frac{z}{2}$

is  parallel to the plane

$2x+2y-z=6$