Why does scalar quantity not have direction

According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$.

The current is defined according to a surface $A_t$ (and is a local quantity: the position of the surface). Since the surface may be tilted (not perpendicular to $\vec{n_d}$), then in general, we should write $$I=n\;q \; \vec{A_t}.\vec{v_d}=n\;q\;A_t \; \vec{n_{A_t}}.\vec{v_d}=n\;q\;A_t \; cos(\theta).v_d$$ where $\theta$ is the angle between $\vec{A_t}$ and $\vec{v_d}$ (If the charge $q$ is negative, $\vec{I}$ and $\vec{n_d}$ would have opposite orientations, which fits with the usual convention of the electric current). Of course, if we have considered a surface $A$ which is perpendicular to the drift, the current would be the same but we would write $$I=n\;q \; \vec{A}.\vec{v_d}=n\;q\;A \; v_d$$

That is, let's talk about the densities. We have $$\mathrm{d}I=n\;q\;v_d\;\mathrm{d}A=n\;q\;v_d\;cos(\theta)\;\mathrm{d}A_t$$ The scalar current density is given by $$j=\frac{\mathrm{d}I}{\mathrm{d}A}=n\;q\;v_d=\frac{1}{cos(\theta)}\frac{\mathrm{d}I}{\mathrm{d}A_t}$$ Yet, you see that you have to be careful about the differential area that you put in the denominator. The current could then be calculated as $$I=\int_A j\;\mathrm{d}A=\int_{A_t} j\;cos(\theta)\;\mathrm{d}A_t$$ Here also, we see a possible source of confusion. This can be fixed if we define a vector current density as $$\vec{j}=j\;\vec{n_d}=n\;q\;v_d\;\vec{n_d}$$ The direction of $\vec{j}$ is resolved too: it's that of the local drift in the considered material position. That direction may be different than that of the average drift of the whole current $\vec{I}$. That is, the expression of the current can be written simply as $$I=\int_A \vec{j}\;\mathrm{d}\vec{A}=\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}$$ Or, you could use your own convention and write $$\vec{I}=\bigg(\int_A \vec{j}\;\mathrm{d}\vec{A}\bigg)\;\vec{n_d}=\bigg(\int_{A_t} \vec{j}\;\mathrm{d}\vec{A_t}\bigg)\;\vec{n_d}$$

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