Edit: Perhaps a little too much background info, for the actual question please scroll down.
I was thinking how to come up with a graphical explanation for the Gram-Schmidt orthogonalization (as it would be more intuitive for people new to the method). However, I ran intro a problem in the case where your vectors span the $\mathbb{R}^3$.
As for the $\mathbb{R}^2$ it's easy:
In order to get $v_2$ from the known vector $v$ and $w_1$, observe that $v = v_1 + v_2$ and hence $v_2 = v – v_1$. Therefore we have to find $v_1$.
Observe that the length of $v_1$ , i.e. $\|v_1\|$, equals $\|v\| \; cos \varphi$. The direction of $v_1$ is the same as the first orthogonal vector $w_1$. So $v_1 = (\|v\| \; cos \varphi) \dfrac{w_1}{\|w_1\|}$.
This looks a little like an inner product, let's write it like one.
$$\begin{align}
v_1 = (\|v\| \; cos \varphi) \dfrac{w_1}{\|w_1\|} = (\|v\| \; cos \varphi) \dfrac{w_1}{\|w_1\|} \dfrac{\|w_1\|}{\|w_1\|} = \dfrac{<v,w_1>}{<w_1,w_1>}w_1
\end{align}$$
And therefore, $v_2 = v – v_1 = v – \dfrac{<v,w_1>}{<w_1,w_1>}w_1$.
Now for $\mathbb{R}^3$, the same idea.
Because $u_3 = v_3 – y$, we have to find $y$. Well, $y = y_1 + y_2$. The length of $y$ equals $\|v_3\| cos \theta$.
Therefore, $y_1 = \|v_3\| cos \theta \; cos \varphi \dfrac{w_1}{\|w_1\|}$, and $y_2 = \|v_3\| cos \theta \; sin \varphi \dfrac{w_2}{\|w_2\|}$.
But from the normal derivation of Gram-Schmidt, $y_1$ should be $\dfrac{<v_3,w_1>}{<w_1,w_1>}w_1$ and $y_2$ should be $\dfrac{<v_3,w_2>}{<w_2,w_2>}w_2$.
So my question, how can I get to this result? Say the angle between $v_3$ and $w_1$ is $\psi$, then from the above equations for $y_1$ it follows that $cos\psi = cos\varphi \; cos \theta$…?
Edit: So the actual question is, why is $cos\psi = cos\varphi \; cos \theta$? Theta $(\theta)$ is the angle between $v_3$ and $y$, phi $(\varphi)$ is the angle between $y$ and $w_1$. Psi $(\psi)$ would be the angle between $v_3$ and $w_1$.
Now I think of it, I could've saved myself and you as reader some time since the actual problem only has to do with angles. Like I said, I know two angles $\varphi$ and $\theta$, how can I get $\psi$ from them such that the two expressions for $y_1$ are the same (and a similar question for $y_2$)?
Best Answer
Well I found a solution, but it involves inner products and splitting inner products. If someone knows how to do it without this, please let me know.
The inner product $<v_3,y>$ can be written as $<v_3,y_1+y_2> = <v_3,y_1> + <v_3,y_2>.$ Furthermore $<v_3,y> = \|v_3\|\|y\| \cos\theta$, $<v_3,y_1> = \|v_3\|\|y_1\| \cos \psi = \|v_3\|\|y\| \cos\varphi \cos \psi$ and $<v_3,y_2> = \|v_3\|\|y_2\| \cos\omega = \|v_3\|\|y\| \sin\varphi \cos \omega$.
Now, divide by $\|v_3\|\|y\|$ such that $\cos \theta = \cos \varphi \cos \psi + \sin \varphi \cos \omega$. Write $\cos \theta$ as $\cos \theta (\cos^2\varphi + \sin^2 \varphi) = \cos \theta \cos^2\varphi + \cos\theta \sin^2\varphi$.
Therefore, $\cos \psi = \cos \theta \cos \varphi$ (and $\cos \omega = \cos \theta \sin \varphi$).