What Vector Spans the Line Defined by These Two Equations?

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Discussion Overview

The discussion revolves around finding a vector that spans the line defined by two linear equations: $x+6y+3z=0$ and $x+3y+4z=0$. Participants explore various methods to approach this problem, including assumptions about the components of the vector and algebraic manipulations of the equations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant suggests setting the y-component of the vector to 1 to simplify the problem, reasoning that it may lead to "nice" numbers due to the coefficient of y being the largest.
  • Another participant describes a method involving subtracting the second equation from the first to eliminate x, leading to a relationship between y and z, ultimately expressing the vector in terms of y.
  • A participant reports finding the vector (-15, 1, 3) and inquires about a systematic method to represent all possible vectors spanning the line.
  • There is a reiteration of the derived vector form, indicating that any multiple of <-15, 1, 3> spans the line.

Areas of Agreement / Disagreement

Participants generally agree on the approach of manipulating the equations to find a spanning vector, but there are multiple methods proposed without a consensus on a single systematic approach.

Contextual Notes

The discussion does not resolve the question of the most systematic method, and participants express varying degrees of confidence in their approaches.

TheFallen018
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Hey,

I've got this problem that I'm trying to work out. I've tried a couple of things, but they don't really get me anywhere.

Here's the problem

Find a vector that spans the line defined by these two equations.
$x+6y+3z=0$
$x+3y+4z=0$

What would be the best way to go about this? Thanks :)
 
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Hi fallen angel,

There are many, many ways to approach this.

The easiest I can think of is the following.
Let's assume that the vector we are searching for has a non-zero y-component.
Then we can set y arbitrarily to 1.
Now solve the system.

Btw, I've picked y to set to 1, since y is the coordinate with the largest coefficient (6), giving us the biggest chance that the other numbers are 'nice' numbers.
 
I like Serena said:
Hi fallen angel,

There are many, many ways to approach this.

The easiest I can think of is the following.
Let's assume that the vector we are searching for has a non-zero y-component.
Then we can set y arbitrarily to 1.
Now solve the system.

Btw, I've picked y to set to 1, since y is the coordinate with the largest coefficient (6), giving us the biggest chance that the other numbers are 'nice' numbers.
Hey, thanks. That makes sense. I ended up coming up with (-15,1,3), which worked nicely. However, I'm curious about other ways to do this. Is there a fairly systematical method that would allow you to come up with a vector that would contain all values? Something that would look like (-15,1,3)+t(x,y,z).

Thanks
 
TheFallen018 said:
Hey,

I've got this problem that I'm trying to work out. I've tried a couple of things, but they don't really get me anywhere.

Here's the problem

Find a vector that spans the line defined by these two equations.
$x+6y+3z=0$
$x+3y+4z=0$

What would be the best way to go about this? Thanks :)
The first thing I see is that the two equations start with "x". If we subtract the second equation from the first, we eliminate "x" and get 3y- z= 0. So z= 3y. Putting 3y in for z the two equations become x+ 6y+ 9y= x+ 15y= 0 and x+ 3y+ 12y= x+ 15y= 0. In either case, x= -15y. So <x, y, z>= <-15y, y, 3y>= y<-15, 1, 3>. <-15, 1, 3>, or any multiple of it, spans that line.
 

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