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?
 A: 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
$$\vec a-\vec b=\vec 0 \implies \vec a=\vec b$$
A: You could also do this using the fact that $\vec{v} \cdot \vec{w} = \vert\vec{v}\rvert \lvert \vec{w} \rvert \cos\theta$ and $\lvert\vec{v} \times \vec{w}\rvert = \vert\vec{v}\rvert \lvert \vec{w} \rvert \sin\theta$. In your case this leads to 
\begin{align*}
(\vec{a} - \vec{b})\ \cdot \ \vec{n} = \lvert\vec{a} - \vec{b}\rvert \lvert\vec{n}\rvert \cos\theta &= 0 \\
\lvert(\vec{a} - \vec{b})\ \times \ \vec{n}\rvert = \lvert\vec{a} - \vec{b}\rvert \lvert\vec{n}\rvert \sin\theta &= 0
\end{align*}
From here, it is safe to cancel the $\lvert\vec{n}\rvert$, as it is unit normal. We cannot safely cancel out the $\cos\theta$ or $\sin\theta$ because these may be $0$, but notice that if $\sin\theta$ = 0, then $\cos\theta \neq 0$ and vice versa. Thus, we are left to the conclusion that $\lvert \vec{a} - \vec{b} \rvert = 0$, so $\vec{a} = \vec{b}$.
A: One definition of the cross product $v , w$ is the unique element $v \times w$ that satisfies
$\langle x , v \times w \rangle = \det \begin{bmatrix} x & v & w \end{bmatrix}$.
Let $d=a-b$. If $d \times n = 0$, then $d,n$ must lie on the same line (otherwise we could
find an $x$ such that the above determinant is non zero). Hence we can write
$d = \lambda n$ for some $\lambda$.
Then $\langle d,n \rangle = \lambda  = 0$. Hence $a=b$.
