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?
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: $$ B' = -\tau N \implies B'\cdot N = -\tau N \cdot N = -\tau. $$