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I've seen a claim which I totally agree with:

All vectors with the same unit vector are parallel.

But I've also seen a claim which is converse to the above:

All parallel vectors have the same unit vector

I don't think it's entirely true which means that either I'm wrong or the claim is false.

Parallel vectors are defined with scalar multiplication. Vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel if one is a scalar multiple of the other:

$$\mathbf{u} \parallel \mathbf{v} \iff \exists c\in \mathbb{R} \, | \, \mathbf{u}=c \mathbf{v}$$

This means that parallel vectors have the same direction ($c>0$) or the opposite direction ($c<0$). An example of the later are two vectors $\mathbf{u}=\langle 1,1 \rangle$ and $\mathbf{v}=\langle -1,-1 \rangle$, i.e. $\mathbf{u}=-\mathbf{v}$. Their unit vectors are $\hat{\mathbf{u}}=\frac{1}{\|\mathbf{u}\|} \langle 1,1 \rangle$ and $\hat{\mathbf{v}}=\frac{1}{\|\mathbf{v}\|} \langle -1,-1 \rangle$, respectively. To me it looks like $\hat{\mathbf{u}} = - \hat{\mathbf{v}}$ which makes sense because unit vector is always in the direction of the original vector and original vectors have opposite directions.

Can you show that the second claim is right for parallel vectors in the opposite direction?

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    $\begingroup$ You are right. Since $\bar{\textbf{u}}$ is not equal to $-\bar{\textbf{u}}$, the second claim is false. $\endgroup$
    – Rai
    Commented Aug 3, 2019 at 13:29

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Sometimes the definition is used that two vectors $u$, $v$ in a real vector space are called parallel if there's a positive $\alpha$ such that $u=\alpha v$. Vectors pointing in the opposite direction (linearly dependent, but with a negative factor) are then called antiparallel.

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