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 slick one-liner proof: $$ \mathbf{a}=(\mathbf{a}\cdot\mathbf{n})\mathbf{n}-(\mathbf{a}\times\mathbf{n})\times\mathbf{n}=\dots=\mathbf{b} $$ – user10354138 Sep 20 at 15:49
  • This is certainly excellent! – nodarkside Sep 20 at 16:10
up vote 6 down vote accepted

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

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

  • Thanks very much @gimusi ! – nodarkside Sep 20 at 15:24
  • @nodarkside You are welcome! Bye – gimusi Sep 20 at 15:25

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 $\vec{a} = \vec{b}$.

  • Thanks very much @Alerra! This proof is also very useful. – nodarkside Sep 20 at 16:02
  • @Alerra you might want to correct your writing: $|\vec v \times \vec w|,$ similarly bellow. – user376343 Sep 20 at 21:07
  • Just fixed it. Although I did it from my phone so it shows up as Community edit I guess? – Alerra Sep 20 at 21:15

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 $a=b$.

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