My Math Forum Intersection of two planes in R3

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 September 30th, 2010, 06:27 PM #1 Newbie   Joined: Sep 2010 Posts: 1 Thanks: 0 Intersection of two planes in R3 I thought I understood equations of planes in R3 and their intersections, but apparently not. I'm very confused by what seems to be a basic problem: find a vector equation for the line of intersection of x + y + z= 0 and x + z = 0. Is x + z= 0 still a plane even though it doesn't have the form Ax + By + Cz = D? I notice that if you set the two equations equal to each other you find that y = 1. Does this mean that the planes intersect on a line where y =1 and all x coordinates are equal to negative z coordinates? Thanks!
October 2nd, 2010, 10:18 AM   #2
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Re: Intersection of two planes in R3

Hello, majami!

Quote:
 Find a vector equation for the line of intersection of $x\,+\,y\,+\,z\:=\:0\text{ and }x\,+\,z\:=\:0$ Is $x\,+\,z\:=\: 0$ still a plane even though it doesn't have the form $Ax\,+\,By\,+\,Cz\:=\:D$?[color=beige] . [/color][color=blue]Yes, it does.[/color] [color=beige]. . [/color][color=blue]It has: A = 1, B = 0, C = 1, D = 0[/color]

$\begin{array}{ccccccc}\text{We have [1]:}=&x\,+\,y\,+\,z=&0 \\ \text{Subtract [2]:}=&x \;\;\;\;\;\;+\,z=&0 \\ \text{and we have:}=&\;\;\;y\;\;\;=&0 \end{array}=$

$\text{From [2], we have: }\;x \:=\:-z$

$\text{We have: }\;\begin{Bmatrix}x=&-z \\ \\ \\ y=&0 \\ \\ \\ z=&z \end{Bmatrix}=$

$\text{On the right, replace }z\text{ with a parameter } -t:\;\;\begin{Bmatrix}x=&t \\ y=&0 \\ z=&-t \end{Bmatrix}=$

$\text{The vector equation is: }\;\vec v \;=\;t\cdot \vec i \,-\,t\cdot\vec k$

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