# Linear Algebra: Norm Help!

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

Staff Emeritus
Gold Member
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?

Staff Emeritus
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.