Linear Algebra: Norm Help!

  • #1

Homework Statement


I'm stuck on this review problem for our final:

The projection of X onto (a,b) = (a,b)
X is orthogonal to (-a,b)
Describe the norm of X in terms of a and b.



The Attempt at a Solution



I drew everything out on a Cartesian system, with the vector X being perpendicular to (-a,b) and then draw the projection of (a,b). I saw that since X is orthogonal to (a,b) it would have a slope of a/b and then attempted to use (y-y1)=m(x-x1) but that failed.
I really have no idea how to solve this and would greatly appreciate any help. Thanks
 

Answers and Replies

  • #2
Tom Mattson
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Forget about writing down an equation of the line that is parallel to [itex]\vec{x}[/itex].

Instead let [itex]\vec{x}=(x_1,x_2)[/itex]. Your goal is to write down a 2 by 2 system of equations in [itex]x_1[/itex] and [itex]x_2[/itex] with coefficients in terms of [itex]a[/itex] and [itex]b[/itex].

The projection of X onto (a,b) = (a,b)
And what does that mean, mathematically? This will give you one of the 2 equations that you need.

X is orthogonal to (-a,b)
Same question: What does that mean, mathematically? This will give you the other of the 2 equations that you need.

Once you have a 2 by 2 system of equations for the components of [itex]\vec{x}[/itex], you can compute its norm.
 
  • #3
Forget about writing down an equation of the line that is parallel to [itex]\vec{x}[/itex].

Instead let [itex]\vec{x}=(x_1,x_2)[/itex]. Your goal is to write down a 2 by 2 system of equations in [itex]x_1[/itex] and [itex]x_2[/itex] with coefficients in terms of [itex]a[/itex] and [itex]b[/itex].



And what does that mean, mathematically? This will give you one of the 2 equations that you need.
Using the projection formula I get [(ax1+bx2)/(a^2+b^2)]*(a,b)=(a,b) so therefore x1=a and x2=b so you get 1*(a,b)=(a,b)


Same question: What does that mean, mathematically? This will give you the other of the 2 equations that you need.
Here I get -ax1+bx2=0 which I rearrange to bx2+ax1

Once you have a 2 by 2 system of equations for the components of [itex]\vec{x}[/itex], you can compute its norm.
Then I get X=SqRt(a^2+b^2) but I don't think this is right.
Can anyone point out where I'm making an error?
 
  • #4
Tom Mattson
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Using the projection formula I get [(ax1+bx2)/(a^2+b^2)]*(a,b)=(a,b)
Yes.

so therefore x1=a and x2=b so you get 1*(a,b)=(a,b)
No. You have to find the solution that satisfies both equations. The solution you wrote down only satisfies one of them.

Here I get -ax1+bx2=0 which I rearrange to bx2+ax1
You mean [itex]bx_2=ax_1[/itex], which is correct. Now solve that equation for either [itex]x_1[/itex] or [itex]x_2[/itex] (your choice), and plug the result into the other equation.
 

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