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?

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    $\begingroup$ I don't think the statement is true. The unit vectors could be pointing in opposite directions. $\endgroup$ – B. Goddard Aug 4 '19 at 12:31
  • $\begingroup$ I thought as you! I even asked a question about it. That's why I restricted $\lambda$ to positive numbers so only the same direction is considered and the statement should be true (if $\lambda$ is negative then yes unit vectors are pointing in opposite directions and aren't equal) $\endgroup$ – oskarryn Aug 4 '19 at 12:36
  • $\begingroup$ @B.Goddard I think the word for that is anti parallel. $\endgroup$ – user572457 Aug 4 '19 at 20:05
  • $\begingroup$ @Random-15 Yes, but they are also called parallel. $\endgroup$ – B. Goddard Aug 4 '19 at 21:02
  • $\begingroup$ @B.Goddard oh! I thought that parallel and anti parallel were different in the vector world. Are sure that they are the same? I have always been taught that they are different, but then, I'm still in high school. $\endgroup$ – user572457 Aug 4 '19 at 21:03

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.


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.

  • $\begingroup$ I only see $\| \mathbf{u}\|$ in terms of $\lambda$, $\mathbf{v}$, and $\hat{\mathbf{u}}$: $\mathbf{u}= \lambda\mathbf{v} \to \| \mathbf{u}\| \hat{\mathbf{u}} = \lambda\mathbf{v} \to \| \mathbf{u}\| = \frac{\lambda \mathbf{v}}{\hat{\mathbf{u}}}$. Similarly I can express $\hat{\mathbf{u}}$ in terms of $\lambda$, $\mathbf{v}$, and $\| \mathbf{u}\|$: $\hat{\mathbf{u}} = \frac{\lambda \mathbf{v}}{\| \mathbf{u}\|}$. Is that what you meant? $\endgroup$ – oskarryn Aug 4 '19 at 14:21

$$\exists \lambda\in \mathbb{R}^+ \, : \, \mathbf{u}= \lambda\mathbf{v} \iff \hat{\mathbf{u}} = \hat{\mathbf{v}}, \text{ where }\mathbf{u},\mathbf{v}\ne \textbf{0}$$

$"\implies"$ 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.

  • $\begingroup$ Thanks for your answer! $\endgroup$ – oskarryn Aug 4 '19 at 14:42
  • $\begingroup$ I see that the initial scalar $\lambda$ modified by lengths of vectors must remain a scalar so unit vectors are parallel to each other (or antiparallel). But I don't see why was that necessary to show. Though, it clicked for me that given $\lambda$ s.t. $\mathbf{u}=\lambda\mathbf{v}$ exists, it must hold that the value of $\lambda$ is equal to $\frac{\|\mathbf{u}\|}{\|\mathbf{v}\|}$ which cancels out lengths and leaves $\hat{\mathbf{u}} = \hat{\mathbf{v}}$ $\endgroup$ – oskarryn Aug 4 '19 at 15:18
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    $\begingroup$ @oskarryn I used this fact because it proves that $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are parallel, and since they are unit vectors, that implies that they are also equal. $\endgroup$ – Mr. Xcoder Aug 4 '19 at 15:22
  • $\begingroup$ Oh wow! I didn't get that vectors coincide means they are equal. All clear now, your answer rocks. Btw. what do you mean exactly by modulus of a vector? $\endgroup$ – oskarryn Aug 4 '19 at 15:31
  • $\begingroup$ Oops, it was just a typo. I meant magnitude, so for a vector $\textbf{u}$, $\|\textbf{u}\|$ would be the magnitude. $\endgroup$ – Mr. Xcoder Aug 4 '19 at 15:33

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