Prove that vectors are parallel iff their unit vectors are equal How to prove that vectors are parallel iff their unit vectors are equal?
$$\mathbf{u} \parallel \mathbf{v} \iff \hat{\mathbf{u}} = \hat{\mathbf{v}}$$
A vector can be written as a scalar multiple of its magnitude and unit vector in its direction: $\mathbf{u}=\|\mathbf{u}\| \hat{\mathbf{u}}$. Intuitively, unit vectors convey the direction and any two vectors with the same unit vector must have the same direction. But how to prove it?
I started:
Vector $\mathbf{u}$ is in the direction of nonzero vector $\mathbf{v}$ iff there exists a positive scalar $\lambda$ which scales vector $\mathbf{v}$ to be equal $\mathbf{u}$ (I don't consider antiparallel vectors here):
$$\mathbf{u} \parallel \mathbf{v} \iff \exists \lambda\in \mathbb{R}^+ \, : \, \mathbf{u}= \lambda\mathbf{v}$$
Hence I try to prove
$$\exists \lambda\in \mathbb{R}^+ \, : \, \mathbf{u}= \lambda\mathbf{v} \iff \hat{\mathbf{u}} = \hat{\mathbf{v}}$$
I'm stuck, any hints?
 A: The question should be rephrased as 
Prove that for  two non zero vectors $u$ and $v$ , $$u=\lambda v \iff \frac {u}{||u||}=\frac {v}{||v||}$$ 
The proof is straightforward. 
A: If $\mathbf{u} = \lambda \mathbf{v}$ for $\lambda > 0$ then you can similarly find $||\mathbf{u}||$ in terms of $\lambda$ and $\mathbf{v}$. Hence, you can express $\hat{\mathbf{u}}$ in terms of $\lambda$ and $\mathbf{v}$.   
For the reverse implication, note $\hat{\mathbf{u}} = \hat{\mathbf{v}}$ implies that $\frac{\mathbf{u}}{||\mathbf{u}||} = \frac{\mathbf{v}}{||\mathbf{v}||},$ which you can manipulate to complete this direction. 
A: 
$$\exists \lambda\in \mathbb{R}^+ \, : \, \mathbf{u}= \lambda\mathbf{v} \iff \hat{\mathbf{u}} = \hat{\mathbf{v}}, \text{ where }\mathbf{u},\mathbf{v}\ne \textbf{0}$$

$"\implies"$  Let $\mathbf{u}=\|\mathbf{u}\| \hat{\mathbf{u}}$ and $\mathbf{v}=\|\mathbf{v}\| \hat{\mathbf{v}}$. Then, the statement $\exists \lambda\in \mathbb{R}^+ \, : \, \mathbf{u}= \lambda\mathbf{v}$ is equivalent to $$\exists \lambda\in \mathbb{R}^+ \, : \,\|\mathbf{u}\| \hat{\mathbf{u}}=\lambda\|\mathbf{v}\| \hat{\mathbf{v}}\iff \hat{\mathbf{u}}=\dfrac{\lambda \|\mathbf{v}\|}{\|\mathbf{u}\|}\hat{\mathbf{v}}$$
Define $\lambda'=\dfrac{\lambda \|\mathbf{v}\|}{\|\mathbf{u}\|}$. This way, $\hat{\mathbf{u}}=\lambda'\hat{\mathbf{v}}$, so the vectors $\hat{\mathbf{u}}, \hat{\mathbf{v}}$ are either parallel or antiparallel, depending on the sign of $\lambda'$, which we'll analyse: $\text{sgn}(\lambda')=\text{sgn}\left(\dfrac{\lambda \|\mathbf{v}\|}{\|\mathbf{u}\|}\right)=\text{sgn}\left(\dfrac{\|\mathbf{v}\|}{\|\mathbf{u}\|}\right),\:\lambda >0$. Now, I think it's just a matter of convention: if one chooses to work with the fact that the modulus of a vector is always positive and that the orientation of the unit vector coincides with the orientation of the vector itself (which I think is the most widely-used convention), then the statement is indeed true (because that means that $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are parallel – and since they are unit vectors, their magnitude is $1$ so they actually coincide). 
$"\:\Longleftarrow\:"$ Again, let $\mathbf{u}=\|\mathbf{u}\| \hat{\mathbf{u}}$ and $\mathbf{v}=\|\mathbf{v}\| \hat{\mathbf{v}}$. Then, the statement $\hat{\mathbf{u}}=\hat{\mathbf{v}}$ is equivalent to
$$\dfrac{\mathbf{u}}{\|\mathbf{u}\|}=\dfrac{\mathbf{v}}{\|\mathbf{v}\|}\iff \mathbf{u} =\dfrac{\|\mathbf{u}\|}{\|\mathbf{v}\|}\mathbf{v}=\lambda\mathbf{v},\text{ where }\lambda=\dfrac{\|\mathbf{u}\|}{\|\mathbf{v}\|}$$
If one uses the aforementioned convention, then $\lambda>0$ so we've completed the proof.
