Suppose I have a vector $\overrightarrow{a}$. If I wanted to describe its magnitude, there's a standard notation for that, $\lVert\overrightarrow{a}\rVert$. Is there a standard notation for its orientation?

I know that I break up the vector into its horizontal and vertical components, I can express the angle as $\theta=\arctan\frac{\lVert\overrightarrow{a_y}\rVert}{\lVert\overrightarrow{a_x}\rVert}$, but the notation is awfully cumbersome to write that way, not to mention that for this to work in all quadrants I’d need to use signed magnitude, almost an oxymoron.

Some have suggested appealing to unit vectors, where $\mathbf{\hat{a}}=\frac{\overrightarrow{a}}{\lVert\overrightarrow{a}\rVert}$, but that’s just kicking the can down the road. It’s fine to say that $\overrightarrow{a}$ and $\mathbf{\hat{a}}$ have the same orientation, but they’re still both vectors, even if the unit vector’s magnitude is $1$; I can’t just write $\mathbf{\hat{a}}=\theta$.

What I’ve been using thus far in my notes is $\overrightarrow{a_\theta}$, figuring that if the $x$ and $y$ components can be written as $\overrightarrow{a_x}$ and $\overrightarrow{a_y}$, then a similar notation could be used to denote the angle component of the vector. I gather that’s not standard notation, however, hence my question.

  • $\begingroup$ Divide the vector by its magnitude to get a unit vector in its direction $\endgroup$ – J. W. Tanner Jul 8 '20 at 22:51

Hat notation is probably what you're after $$\hat{a} = \frac{\vec{a}}{\|\vec{a}\|} .$$

  • $\begingroup$ How does this help? I’ve defined a unit vector with the same orientation. Now what? I wouldn’t write $\hat{a}=\theta$, since $\hat{a}$ is still a vector just like the one I started with, with magnitude as well as orientation. $\endgroup$ – DonielF Jul 8 '20 at 23:09
  • 2
    $\begingroup$ There is a one-to-one correspondence between each a-hat and the angles in $[0,2\pi )$. You can talk about the a-hat as if they were particular angles in that way. $\endgroup$ – Fede Poncio Jul 8 '20 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.