Two vectors with the same normal surface projection and the same normal surface cross product, are equal? I have two vectors, $\mathbf a$ and $\mathbf b$, that fulfill the following conditions: $(\mathbf a-\mathbf b)\cdot \mathbf n= 0$ $(\mathbf a-\mathbf b)\times \mathbf n=\mathbf 0$ being $\mathbf n$ a unit surface normal. My question is, is $\mathbf a = \mathbf b$ ? I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?

Yes it is correct, indeed we have that

* $(\vec a-\vec b)\times \vec n=\vec 0 \implies \vec a-\vec b$ is a multiple of $\vec n$ that is $\vec a-\vec b=k\vec n$
* $(\vec a-\vec b)\cdot \vec n=0 \implies \vec a-\vec b$ is orthogonal to $\vec n$ that is $k\vec n\cdot \vec n=0 \implies k=0$



therefore

$$\vec a-\vec b=\vec 0 \implies \vec a=\vec b$$