Sign of angle between vectors

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Say I have a 2D vector $\vec{AB}$ connecting two points

and

. The midpoint of the vector,

, is connected to another point

, which is not collinear with

and

.

If I rotate $\vec{AB}$ by an angle $\theta$ , where $\theta<\pi$ , then $\vec{AB}$ has the same direction as $\vec{MC}$ . (Similarly, if I rotate $\vec{AB}$ in the other direction by an angle $(2\pi-\theta)$ then $\vec{AB}$ has the same direction as $\vec{MC}$ .)

I can figure out $\theta$ using the dot product of $\vec{MC}$ and $\vec{MB}$ . My question is, how do I figure out if $\theta$ is in the clockwise or counterclockwise direction?

Shadow · **Posted:** Fri, 6 Jan 2012 21:44:53 UTC

DannyBoy wrote:

Say I have a 2D vector $\vec{AB}$ connecting two points

and

. The midpoint of the vector,

, is connected to another point

, which is not collinear with

and

.

If I rotate $\vec{AB}$ by an angle $\theta$ , where $\theta<\pi$ , then $\vec{AB}$ has the same direction as $\vec{MC}$ . (Similarly, if I rotate $\vec{AB}$ in the other direction by an angle $(2\pi-\theta)$ then $\vec{AB}$ has the same direction as $\vec{MC}$ .)

I can figure out $\theta$ using the dot product of $\vec{MC}$ and $\vec{MB}$ . My question is, how do I figure out if $\theta$ is in the clockwise or counterclockwise direction?

A priori you cannot tell, because it will depend on an orientation, which you haven't chosen.

If you are using the standard orientation, you should be able to convince yourself that you STILL cannot tell because cosine is an even function, so the dot product doesn't care if it's positive or negative.

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Thanks for the reply, Shadow! I guess that's my point: the dot product cannot tell me the direction, so how do I find out?

Shadow · **Posted:** Fri, 6 Jan 2012 22:06:36 UTC

DannyBoy wrote:

Thanks for the reply, Shadow! I guess that's my point: the dot product cannot tell me the direction, so how do I find out?

If all you know is the dot product, you cannot find out. Even if you know one of the vectors you cannot figure that out, because all the dot product does is tell you how much of one vector is in the direction of the other, so you can get the direction of the other vector if you know one and you know their dot product, but that's about it.

To see this note if one vector has components (a,b)

known and the other has components (x,y)

unknown and you have their dot product is c, then all you know is ax+by=c

, i.e. what line your other vector lies on.

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Thanks again for the reply, Shadow. Perhaps I'm not articulating myself very well: I know both vectors.

In short, if I have two vectors, how do I determine which direction to rotate one of these vectors such that it points the same way as the other vector providing I rotate it through the smallest angle possible?

Shadow · **Posted:** Fri, 6 Jan 2012 22:32:22 UTC

DannyBoy wrote:

Thanks again for the reply, Shadow. Perhaps I'm not articulating myself very well: I know both vectors.

In short, if I have two vectors, how do I determine which direction to rotate one of these vectors such that it points the same way as the other vector providing I rotate it through the smallest angle possible?

Oh, that's easy. Just look at the orientation between them, if C is to the right of the line through A and B, then go clockwise, if not go counterclockwise, draw a picture of both situations and you should be able to see why this is.

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

But how do I do it mathematically? I have around half a million cases to consider, so would prefer not to draw pictures for each!

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Okay, here's my attempted solution. Assuming two 2D vectors, $\vec{a}$ and $\vec{b}$ , the direction of smallest necessary rotation needed for $\vec{a}$ to align with $\vec{b}$ can be determined as follows.

$\alpha=\arctan2(y_b,x_b)-\arctan2(y_a,x_a)$

Counterclockwise if $\pi>\alpha>0$ or $\alpha<-\pi$
Clockwise if $-\pi<\alpha<0$ or $\alpha>\pi$
Neither if $\alpha=0$
Either if $\alpha=\pi$ or $\alpha=-\pi$

Shadow · **Posted:** Mon, 9 Jan 2012 20:15:20 UTC

DannyBoy wrote:

But how do I do it mathematically? I have around half a million cases to consider, so would prefer not to draw pictures for each!

You said you know the vectors, look at their coordinates.

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Is there a more efficient way of determining the direction from the coordinates than my solution, which seems rather clunky?

Shadow · **Posted:** Mon, 9 Jan 2012 20:47:48 UTC

Probably, but I personally don't see why, you asked how to do it, and that works, so why rock the boat?

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Because I have to perform the operations around half a million times in an R program, so the more efficient the solution, the faster my code will run.

outermeasure · **Posted:** Tue, 10 Jan 2012 14:18:43 UTC

DannyBoy wrote:

Because I have to perform the operations around half a million times in an R program, so the more efficient the solution, the faster my code will run.

You need only the sign of x_1y_2-x_2y_1

, where $(x_1,y_1)=\overrightarrow{AB},\quad (x_2,y_2)=\overrightarrow{MC}$ . However, note the subtraction of possibly nearly equal quantity.

DannyBoy · **Joined:** Mon, 2 May 2005 19:58:46 UTC **Posts:** 86

Wow, very efficient! Thanks, outermeasure! So, your approach looks like the determinant of $\left[\begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \\ \end{array}\right]$ . Could you explain why it works or point me in the direction of a webpage?

outermeasure · **Posted:** Tue, 10 Jan 2012 15:29:50 UTC

DannyBoy wrote:

Wow, very efficient! Thanks, outermeasure! So, your approach looks like the determinant of $\left[\begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \\ \end{array}\right]$ . Could you explain why it works or point me in the direction of a webpage?

You should know the signed area of triangle OPQ

is $\dfrac{1}{2}\det\begin{pmatrix}x_P & y_P\\ x_Q & y_Q\end{pmatrix}$ (assuming usual orientation).

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Sign of angle between vectors

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