Problem with the proof of of $\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta$ in Simon, Blume 1994 In Simon, Blume 1994, p. 215, we find the following proof of the identity
$$
\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta
$$

Without loss of generality, we can work with $\mathbf{u}$ and $\mathbf{v}$ as vectors with tails at the origin $\mathbf{0} ;$ say $\mathbf{u}=\overrightarrow{O P}$ and $v=\overrightarrow{O Q}$. Let $\ell$ be the line through the vector $\mathbf{v}$, that is, the line through the points 0 and $Q$. Draw the perpendicular line segment $m$ from the point $P$ (the head of $\mathbf{u}$ ) to the line $\ell$, as in Figure 10.20 . Let $R$ be the point where $m$ meets $\ell$. Since $R$ lies on $\ell, \overrightarrow{O R}$ is a scalar multiple of $\mathbf{v}=\overrightarrow{O Q}$. Write $\overrightarrow{O R}=t v .$ Since $\mathbf{u}, t \mathbf{v},$ and the segment $m$ are the three sides of the right triangle $O P R,$ we can write $m$ as the vector $\mathbf{u}-t \mathbf{v}$. Since $\mathbf{u}$ is the hypotenuse of this right triangle,
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
On the other hand, by the Pythagorean Theorem and Theorem $10.2,$ the square of the length of the hypotenuse is:
$\|\mathbf{u}\|^{2}=\|t \mathbf{v}\|^{2}+\|\mathbf{u}-t \mathbf{v}\|^{2}$
$$
\begin{array}{l}
=t^{2}\|\mathbf{v}\|^{2}+(\mathbf{u}-t \mathbf{v}) \cdot(\mathbf{u}-t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\mathbf{u} \cdot \mathbf{u}-2 \mathbf{u} \cdot(t \mathbf{v})+(t \mathbf{v}) \cdot(t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\|\mathbf{u}\|^{2}-2 t(\mathbf{u} \cdot \mathbf{v})+t^{2}\|\mathbf{v}\|^{2}
\end{array}
$$
or
$$
2 t(\mathbf{u} \cdot \mathbf{v})=2 t^{2}\|\mathbf{v}\|^{2}
$$
It follows that
$$
t=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^{2}}
$$
Plugging equation (5) into equation (4) yields
$$
\cos \theta=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$

This identity is then used to prove the following theorem

Theorem 10.4 The angle between vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbf{R}^{\mathbf{n}}$ is
(a) acute, if $\mathbf{u} \cdot \mathbf{v}>0$
(b) obtuse, if $\mathbf{u} \cdot \mathbf{v}<0$,
(c) right, if $\mathbf{u} \cdot \mathbf{v}=0$.

However, I don't understand this proof fully since it seems that
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{|t|\|\mathbf{v}\|}{\|\mathbf{u}\|}\neq\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
Especially, this would imply that
$$
\cos \theta=\frac{|\mathbf{u} \cdot \mathbf{v}|}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$
which would contradict the theorem 10.4. Am I missing something, or is there an error in this proof?
Thank you in advance for your help!

The quoted proof was careless at one point. The generally correct expression is $t\Vert v\Vert/\Vert u\Vert$ rather than $|t|\Vert v\Vert/\Vert u\Vert$. The sign of $t$ boils down to whether the multiple of $v$ onto which $u$ is projected is parallel or antiparallel to $v$.
