# Thread: [proof] Angles Between Two Vectors and Their Projections Are Equal

1. ## [proof] Angles Between Two Vectors and Their Projections Are Equal

Q: Assume $\vec{x}, \vec{y} \in\mathbb{R}^n$ such that $\vec{x}\cdot\vec{y}\neq0$, prove that the angle between $\vec{x}$ and $\vec{y}$ is equal to the angle between $proj_{\vec{x}}\vec{y}$ and $proj_{\vec{y}}\vec{x}$

I've tried various things but I have yet to discover a good (formal) way of tackling this. I've written on the first line:

$\cos\theta=\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\ |\vec{y}\|}=\frac{proj_{\vec{x}}\vec{y}\cdot proj_{\vec{y}}\vec{x}}{\|proj_{\vec{x}}\vec{y}\|\| proj_{\vec{y}}\vec{y}\|}$

Now I've tried expanding this but I don't think it leads me anywhere so next I write the definition of projection vector:

$proj_{\vec{x}}\vec{y} = (\frac{\vec{x}\cdot\vec{y}}{\vec{x}\cdot\vec{x}})\ vec{x}$

Here it occurs to me that $\vec{x}\cdot\vec{y}$ can be either positive or negative, therefore the projections will be either in the same direction or the opposite direction of $\vec{x}$ and $\vec{y}$. So this leads me to believe that the projections do have the same angle by the congruence law of intersecting lines(?) Is this somewhat correct? Does it hold for $\mathbb{R}^n$, n>2? How do I format this into a formal proof?

2. The projection of $\vec{x}$ on $\vec{y}$ is, by definition, a vector in the same direction as $\vec{y}$ and the projection of $\vec{y}$ on $\vec{x}$ is a vector in the same direction as $\vec{x}$, so of course, the angle between them is the same.

3. You should know that $\dfrac{\vec{x}\cdot\vec{y} }{\vec{x}\cdot\vec{x}}~\&~\dfrac{\vec{x}\cdot\vec{ y} }{\vec{y}\cdot\vec{y}}$ must have the same sign.

Also $\left\|\text{proj}_{\vec{x}}\vec{y}\right\|=\left| \frac{\vec{x}\cdot\vec{y} }{\vec{x}\cdot\vec{x}} \right|\left\|\vec{x}\right\|$.