Vector that spans two equations

[TheFallen018]

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 :)

 

[I like Serena]

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.

 

[TheFallen018]

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

 

[HOI]

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.