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How do you state that two vectors $\vec{A}$ and $\vec{B}$ have the same direction? I know the symbol $| |$ shows that they are parallel, but is there a symbol like this that shows that the direction of the vectors are equal. I think you would call the two vectors collinear. Is there a symbol to show that two vectors are collinear?

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    $\begingroup$ Is there a reason parallel isn't enough? Because that literally means their directions are exactly the same $\endgroup$
    – Triatticus
    Oct 18, 2017 at 14:20
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    $\begingroup$ I’m not sure if there’s a symbol for linear dependence (but there may well be one), but to show it you can write $\vec{A}=k\vec{B}$ for some (scalar) constant $k$ $\endgroup$ Oct 18, 2017 at 14:21
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    $\begingroup$ Generally the word parallel is used for vectors pointing in the same direction, and anti-parallel is used for vectors which are pointing in opposite directions. $\endgroup$
    – user275377
    Oct 18, 2017 at 14:22
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    $\begingroup$ Perhaps he means one vector is a nonnegative scalar multiple of the other. That is: the case of equality for the triangle inequality. $\endgroup$
    – GEdgar
    Oct 18, 2017 at 14:38
  • $\begingroup$ @aidangallagher4 yeah that's what I ended up using, I was just wondering if there was a better symbol for it. $\endgroup$
    – Bhaskar
    Oct 19, 2017 at 2:40

2 Answers 2

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Not sure if this is considered necrobumping, but my higher level calculus professors in university always used // to notate this. I don't see it anywhere else in the math realm, however.

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“Anti-parallel” definitely means that two (Euclidean) vectors are in opposite directions, whereas “parallel” may just mean that they are collinear (as such, the zero vector can be said to be parallel to all vectors).

To unambiguously describe two vectors as in fact being in the same direction, I'd state that they are a positive scalar multiple of each other.

P.S. Notice though that two vector being collinear does not imply that they are a scalar multiple of each other: e.g., $\mathbf 0$ and $\mathbf k.$

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