How can I find the norm of X in terms of a and b when X is orthogonal to (-a,b)?

[himynameismar]

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

 

[quantumdude]

Forget about writing down an equation of the line that is parallel to $\vec{x}$.

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

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 $\vec{x}$, you can compute its norm.

 

[himynameismar]

Forget about writing down an equation of the line that is parallel to $\vec{x}$.

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



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 $\vec{x}$, 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?

 

[quantumdude]

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 $bx_2=ax_1$, which is correct. Now solve that equation for either $x_1$ or $x_2$ (your choice), and plug the result into the other equation.