is angle a scalar or a vector quantity

- anonymous

is angle a scalar or a vector quantity

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- anonymous

Angle has magnitude only so it is a SCALAR.

- anonymous

Vector quantities have both magnitude and direction. So, angle is a scalar quantity.

- Kainui

If you want to get technical, a vector has a direction and magnitude. But an angle isn't a magnitude, it's a direction. @Emineyy

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- theEric

I found this:
Scalars are quantities that are fully described by a magnitude (or numerical value) alone.
Vectors are quantities that are fully described by both a magnitude and a direction.
http://www.physicsclassroom.com/Class/1DKin/U1L1b.cfm

- theEric

Now, displacement is considered a vector, even in one dimension. Angles are quantities that can have direction, but they don't have to necessarily. There is, for example, \(-270^\circ\). But \(-270^\circ=90^\circ\). So the angle can be described with only a magnitude.
However, the sign might be very important in some equations.
Like, for two-dimensional polar coordinates, you have an \(r(\theta)\), and you can have a curve that is \(r(\theta)=\theta+\theta^2\).
http://www.wolframalpha.com/input/?i=polar+theta%2Btheta^2
Now, direction of the value \(\theta\) is important, and is the measured angle from the \(+x\)-axis. And \(r(-270^\circ)=r\left(-\dfrac{3\pi}{4}\right)=-\dfrac{3\pi}{4}+\left(-\dfrac{3\pi}{4}\right)^2\approx3.19\)
while \(r(90^\circ)=r\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+\left(\dfrac{\pi}{4}\right)^2\approx 1.40\)
That is a function that uses an angle where the sign of the angle is important.

- theEric

But I don't see how angles alone would fill a multidimensional vector that would reflect one angle. I don't see how you can break an angle into a sum unit vectors, and such.
Good luck! Maybe the math section would help! But think about this.
Johnny walks ten feet forward, and five feet backward.
Distance: 15 feet
Displacement: 5 feet
Johnny turns \(90^\circ\) counter clockwise, and \(45^\circ\) clockwise.
Degrees turned, regardless of direction: \(90^\circ +45^\circ = 135^\circ\)
Net angle change: \(90^\circ + (-45^\circ)=45^\circ\).

- theEric

But many times we don't consider direction, unless we have a coordinate system.
So I'm really not sure what angles would be considered.

- anonymous

:)

- theEric

Haha! :)

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- Shane_B

@Kanui: What makes you think an angle has a direction and not a magnitude?

- Kainui

@Shane_B
Well for example, 45 degrees is like saying North East. If someone says my house is north east of here, that doesn't tell you how far, only a direction. There's no magnitude involved since angles are completely circular.
See, if you were to say 45 degrees was a magnitude, then I could say I have something of much greater magnitude, called 405 degrees. But wait, 405-360=45. You're just spinning in circles, you're not talking about magnitudes here.

- Shane_B

I think we'll end up disagreeing here.
North-east is note an angle...it's a direction. Yes, you may have to turn 45 degrees to face North East but that does not mean 45 degrees is a direction. It's a relative magnitude.
By definition, angles are the space between two intersecting lines or surfaces at or close to the point where they meet. It's a magnitude that's typically measured in degrees or radians.

- Kainui

It's a direction relative to your set coordinate axis, not a magnitude, sorry.

- Shane_B

I think you're confusing a bearing with an angle. Wiki "Angle" :)

- theEric

I think it can indicate direction in a coordinate system or indicate angle between two directions without one. But \(25^\circ\neq-25^\circ\) when you consider a coordinate system, so direction matters (as it does with vectors). Here, \(25^\circ\) would be the magnitude, which is the other feature of vectors.
I still have no definitive answer as to what an angle is formally considered. I think all three-dimensional angles can be covered with two components, as is done in spherical coordinates. But I haven't thought about whether the vector math would apply to this two-dimensional angle vector concept. It seems plausible, at first thought.... But then the cross-product (a.k.a. vector product) would require a third angle dimension, requiring the coordinate system to be four-dimensional.

- theEric

I used Wikipedia to find an article on angles. It reads
"Although the definition of the measurement of an angle does not support the concept of a negative angle, it is frequently useful to impose a convention that allows positive and negative angular values to represent orientations and/or rotations in opposite directions relative to some reference."
This means that there are no negative angles and thus there is no direction.

- theEric

Alright... Here's what I just did in my classical mechanics class....
In planar polar coordinates:
\(\hat{e_\theta}=-\hat i \sin(\theta)+\hat j\cos(\theta)\)
where \(\hat {e_\theta}\) is the unit vector in the direction of \(\vec\theta\).
This is used to differentiate \(\vec r\) with respect to time \(t\) to get \(\vec v\), and I would think that I'm being taught widely accepted material, so it seems that the \(\theta\) variable is the magnitude of \(\vec\theta\).
That said, I don't understand how the value for \(\hat{e_\theta}\) was chosen, yet. I'll have to get to that...
____________________________
This is my thoughts, now. Not material from class!
:
This doesn't mean that concept of an \(\sf angle\) is a vector, though. But it seems to me that we use a concept similar to an angle. We look at it differently because
1. We accept that there is direction in angle due to an imposed coordinate system. That gives the concept of direction, with negative and positive.
2. We accept that, as we go around from the \(x\) axis, we don't use the smallest angle (which means the angle would always be in the range \([0,\ 180^\circ]\), or \([-180^\circ,\ 180^\circ]\) with direction relevant).
Putting these two ideas with the idea of an angle, we now have another concept that I still consider to describe angles as I've been using them. The angle has direction in an imposed coordinate system and can be anywhere in the range \((-\infty,\ \infty)\).

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