Two questions about the definition of torsion of a curve The following is from the Wikipedia article on the torsion of a curve:
[The torsion] is found from the equation
$$B'=-\tau N$$
which means
$$\tau =- N \cdot B'$$

Here $N$ is the normal unit vector, and $B$ the binormal unit vector.

$Q_1$: considering that the torsion $\tau$ is a scalar, the first equation implies that $B'$ is parallel to $N$. Why is this the case?
$Q_2$: how does the first equation imply the second?
 A: I feel like this should be a duplicate, but I couldn't find an answer that clearly proves that $B'$ is parallel to $N$. The wikipedia proof also doesn't really go into this. So here goes, with all the details:
We have
$$
B' = (T\times N)' = T'\times N + T\times N'.
$$
The first term gives
$$
T'\times N = \kappa N\times N = 0
$$
(where $\kappa$ is the curvature, though that is irrelevant for us). What about $N'$? We will write $N'$ as a linear combination of $T$, $N$ and $B$.  Since $N$ is of constant length, $N'$ has no component in the $N$ direction. (Proof: $N'\cdot N = \frac12(N\cdot N)' = \frac12(1)'=0$.) So we have $N'=\alpha T + \beta B$ for some scalars $\alpha$ and $\beta$. But then
$$
B' = T\times N' = \alpha T\times T + \beta T\times B = -\beta N.
$$
And we define $\tau=\beta$. Incidentally, $\alpha=-\kappa$, which follows by calculating $(T\cdot N)'$.

Alternatively: Write $B' = \alpha T + \beta N + \gamma B$. We have
$$
0 = (B\cdot T)' = B'\cdot T + B\cdot T' = B'\cdot T + B\cdot (\kappa N) = B'\cdot T 
\implies \alpha = 0,
$$
and
$$
0 = (B\cdot B)' = 2B'\cdot B \implies \gamma = 0.
$$
So
$$
B' = \beta N,
$$
and we define $\tau = -\beta$.

Oh, I forgot your second question, which I answered in the comment:
$$
B' = -\tau N \implies B'\cdot N = -\tau N \cdot N = -\tau.
$$
