# Math Help - Orthogonality

1. ## Orthogonality

Let $\vec{x}$ and $\vec{y}$ be linearly independent vectors in $R^2$. If || $\vec{x}$|| = 2 and ||[ $\vec{y}$|| = 3, what, if anything, can we conclude about the possible values of | $\vec{x}^{T}\vec{y}$|?

So this is what I have so far...

$2 = \sqrt{x_{1}^{2}+x_{2}^{2}}$
$3 = \sqrt{y_{1}^{2}+y_{2}^{2}}$

After some algebra I got...

$\vec{x}^{T}\vec{y}=(\sqrt{4-x_{2}^{2}}\sqrt{4-x_{1}^{2}})(\sqrt{9-y_{2}^{2}}\sqrt{9-y_{1}^{2}})^{T}$

This is where I'm stuck, what can I determine from this? Thanks in advance!

2. Originally Posted by pakman
Let $\vec{x}$ and $\vec{y}$ be linearly independent vectors in $R^2$. If || $\vec{x}$|| = 2 and ||[ $\vec{y}$|| = 3, what, if anything, can we conclude about the possible values of | $\vec{x}^{T}\vec{y}$|?

So this is what I have so far...

$2 = \sqrt{x_{1}^{2}+x_{2}^{2}}$
$3 = \sqrt{y_{1}^{2}+y_{2}^{2}}$

After some algebra I got...

$\vec{x}^{T}\vec{y}=(\sqrt{4-x_{2}^{2}}\sqrt{4-x_{1}^{2}})(\sqrt{9-y_{2}^{2}}\sqrt{9-y_{1}^{2}})^{T}$

This is where I'm stuck, what can I determine from this? Thanks in advance!
If I'm not mistaken this looks like it's an Euclidean dot product, so
$|\vec{x}^T \vec{y}| = ||\vec{x}||~||\vec{y}||~|cos(\theta)|$ which ranges from 0 to 2 * 3 = 6.

-Dan

3. Originally Posted by topsquark
If I'm not mistaken this looks like it's an Euclidean dot product, so
$|\vec{x}^T \vec{y}| = ||\vec{x}||~||\vec{y}||~|cos(\theta)|$ which ranges from 0 to 2 * 3 = 6.

-Dan
How do you figure that it ranges from 0 to 6? I am a bit confused...

4. Originally Posted by pakman
How do you figure that it ranges from 0 to 6? I am a bit confused...
$||\vec{x}||$ and $||\vec{y}||$ are constants. And $|cos(\theta)|$ ranges from 0 to 1.

-Dan

5. Originally Posted by topsquark
$||\vec{x}||$ and $||\vec{y}||$ are constants. And $|cos(\theta)|$ ranges from 0 to 1.

-Dan
Ahh okay makes sense now thanks!