Questions on vectors

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  • #1
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I just learnt vectors as a prerequisite to learning mechanics and I have a few doubts:-

1) For 1D motion, there are only two possible directions, so they can be represented by just a + or - sign, where you have to define which direction is positive and which is negative.

But for 2D how do you represent the direction of a vector? Clearly, just + or - will not do the job. I thought of a few ways to completely describe the direction of a vector in 2D:-

  • Using words such as North, North East, left, right etc. but this isn't precise and will not suffice for vectors that aren't in the 8 simple directions.
  • Defining a line like an x-axis, and using the angle the vector makes with the defined line to describe it's direction.
  • Defining x and y axes and then resolving the vectors along the axes. So the vector will be expressed as the sum of an x-component and a y-component with + and - signs representing the directions of the components along their respective axis.

So what method do people use to describe direction of vectors in 2D?

2) When we are drawing displacement-time graphs, along the displacement axis there is only + and -. So does this mean that we can draw displacement-time graphs for motion in 1D?
 

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  • #2
vanhees71
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Vectors in an n-dimensional vector space can always be represented with respect to a basis. This can be any set of n linearly independent vectors. Then each vector is represented by n (in our case real) numbers via the unique decomposition of the vector with respect to this basis
[tex]\vec{v}=v_1 \vec{e}_1+v_2 \vec{e}_2+\cdots + v_n \vec{e}_n.[/tex]
In mechanics you usually have Euclidean vector spaces, and thus you should choose the basis as a Cartesian system, i.e., you choose vectors of unit length which are perpendicular to each other.

The direction is given by a unit vector
[tex]\vec{n}=n_1 \vec{e}_1 + \cdots + n_n \vec{e}_n.[/tex]
The unit length of the vector is given in terms of Cartesian components by
[tex]|\vec{n}|=\sqrt{n_1^2+n_2^2+\cdots n_n^2}=1.[/tex]

In two dimensions you can give any unit vector in terms of the angle with respect to [itex]\vec{e}_1[/itex] via
[tex]\vec{n}(\varphi)=\cos \varphi \vec{e}_1 + \sin \varphi \vec{e}_2.[/tex]
To see this, just draw this vector, and you can read off its components in terms of the angle using elementary trigonometric properties of rectangular triangles.
 
  • #3
SteamKing
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All three methods are used:

1. The bearing form is used a lot for survey work. In the bearing form, the cardinal directions (NESW) are defined thus: N = +y; E = +x; S = -y; W = -x. Positive angles are oriented counter clockwise. A bearing angle would be specified like this: 30 degrees E of N which would form an angle of 60 degrees with the +x axis.
2. The Magnitude-Angle form would be something like this: 40 N < 45 degrees.
3. The component form, expressed either as an ordered pair (x,y) or as a vector sum (x*i-cap + y*j-cap), where i-cap and j-cap represent the unit vectors in the x and y directions. The component form is used extensively since two vectors can be readily added and vector products (dot and cross) are defined using components.


I don't think you can have a 1D vector.
 
  • #5
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Thanks you so much for your replies! :biggrin:

Vectors in an n-dimensional vector space can always be represented with respect to a basis. This can be any set of n linearly independent vectors. Then each vector is represented by n (in our case real) numbers via the unique decomposition of the vector with respect to this basis
[tex]\vec{v}=v_1 \vec{e}_1+v_2 \vec{e}_2+\cdots + v_n \vec{e}_n.[/tex]
n-dimensions :confused: n can only be 1,2 or 3 right? Also, I have no idea what you mean when you say vector space or linearly independent or what basis you are referring to. But I can always guess :tongue2:

  • vector space I suppose this means the dimensions your vectors are in. So if you have vectors on a paper then you would be talking about a 2D vector space?
  • linearly independent Does this mean they are perpendicular to each other? And it's called independent because you can't resolve any one of the vectors such that they have a component in the same direction as the other vector?
  • basis Does this refer to a set of axes that you are using for reference? In which case for 2D the x and y axes would be the 2 linearly independent vectors that you were talking about?

Also, when you say e1, e2 are you talking about unit vectors? I've not learnt unit vectors but when steamking mentioned them I looked it up and I think that's what you mean. Not sure though :frown: And when you say n (in our case real) numbers I think you meant vectors right?

In mechanics you usually have Euclidean vector spaces, and thus you should choose the basis as a Cartesian system, i.e., you choose vectors of unit length which are perpendicular to each other.
Euclidean vector spaces Whazzat?

The direction is given by a unit vector
[tex]\vec{n}=n_1 \vec{e}_1 + \cdots + n_n \vec{e}_n.[/tex]
The unit length of the vector is given in terms of Cartesian components by
[tex]|\vec{n}|=\sqrt{n_1^2+n_2^2+\cdots n_n^2}=1.[/tex]
Unit length is the same thing as length or magnitude right?

In two dimensions you can give any unit vector in terms of the angle with respect to [itex]\vec{e}_1[/itex] via
[tex]\vec{n}(\varphi)=\cos \varphi \vec{e}_1 + \sin \varphi \vec{e}_2.[/tex]
To see this, just draw this vector, and you can read off its components in terms of the angle using elementary trigonometric properties of rectangular triangles.
So this means you can use the angle form (my 2nd bullet point) to represent a vector because from it you can calculate the components form using simple decomposition (the formula you gave)?

Could you please answer the second question?

All three methods are used:

1. The bearing form is used a lot for survey work. In the bearing form, the cardinal directions (NESW) are defined thus: N = +y; E = +x; S = -y; W = -x. Positive angles are oriented counter clockwise. A bearing angle would be specified like this: 30 degrees E of N which would form an angle of 60 degrees with the +x axis.
2. The Magnitude-Angle form would be something like this: 40 N < 45 degrees.
3. The component form, expressed either as an ordered pair (x,y) or as a vector sum (x*i-cap + y*j-cap), where i-cap and j-cap represent the unit vectors in the x and y directions. The component form is used extensively since two vectors can be readily added and vector products (dot and cross) are defined using components.


I don't think you can have a 1D vector.
What is 40 N < 45 degrees. What is the < indicate? :confused:

One can have one dimensional or even zero dimensional vector spaces. I agree that their practical usefulness is questionable.
Yeah that's what I thought too. But when we talk about motion in a straight line (what I'm doing right now) we are using 1D vectors aren't we? So they are useful.

Thanks again everyone!:smile:
 
  • #6
WannabeNewton
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I assume you haven't take a basic course in linear algebra yet if you haven't encountered terms like basis or linear independence. As such, don't worry about n dimensional vector spaces or even n dimensional Euclidean spaces; you will encounter those terms eventually. Focusing on just ##\mathbb{R}^{2}##, if ##v\in \mathbb{R}^{2}## is a vector then we can write it as ##v = v^{1}e_1 + v^{2}e_{2}## where ##e_{1} = (1,0)^{T}##,##e_{2} = (0,1)^{T}## and ##v^{1},v^{2}\in \mathbb{R}##. Note that ##v\cdot e_{1} = \left \| v \right \|cos\theta = v^{1}## and ##v\cdot e_{2} = \left \| v \right \|\cos(90 - \theta) = \left \| v \right \|\sin\theta = v^{2}## where ##\theta## is the angle above the horizontal. Hence ##\tan\theta = \frac{v^2}{v^1}## so the direction of the vector is simply given by ##\theta = \arctan(\frac{v^2}{v^1})##.

The dimensionality of a vector space can be any natural (a zero dimensional vector space is just the trivial space ##{0}##) or it can be infinite.
 
  • #7
vanhees71
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Be careful the use of tan here! Better is the formula
[tex]\theta=\mathrm{sign}(\nu^2+0^+) \arccos \left (\frac{\nu^1}{\sqrt{(\nu^1)^2+(\nu^2)^2}} \right).[/tex]
This gives an angle [itex]\Theta \in ]-\pi,\pi].[/itex]
 
  • #8
jbriggs444
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[*]linearly independent Does this mean they are perpendicular to each other? And it's called independent because you can't resolve any one of the vectors such that they have a component in the same direction as the other vector?
A set of vectors is linearly independent if there is no element of the set that can be computed as a linear combination of the other elements in the set.

So the set containing (1,2) and (2,4) would not be linearly independent because (2,4) = 2 * (1,2).

But the set containing (1,2) and (1,3) would be linearly independent even though those two vectors are not perpendicular.

In two dimensions, you obviously cannot have a set containing more than two vectors that are linearly independent. Roughly speaking, that is what it means to say that a vector space has two dimensions.
 
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  • #9
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I assume you haven't take a basic course in linear algebra yet if you haven't encountered terms like basis or linear independence. As such, don't worry about n dimensional vector spaces or even n dimensional Euclidean spaces; you will encounter those terms eventually. Focusing on just ##\mathbb{R}^{2}##, if ##v\in \mathbb{R}^{2}## is a vector then we can write it as ##v = v^{1}e_1 + v^{2}e_{2}## where ##e_{1} = (1,0)^{T}##,##e_{2} = (0,1)^{T}## and ##v^{1},v^{2}\in \mathbb{R}##. Note that ##v\cdot e_{1} = \left \| v \right \|cos\theta = v^{1}## and ##v\cdot e_{2} = \left \| v \right \|\cos(90 - \theta) = \left \| v \right \|\sin\theta = v^{2}## where ##\theta## is the angle above the horizontal. Hence ##\tan\theta = \frac{v^2}{v^1}## so the direction of the vector is simply given by ##\theta = \arctan(\frac{v^2}{v^1})##.

The dimensionality of a vector space can be any natural (a zero dimensional vector space is just the trivial space ##{0}##) or it can be infinite.
I'm just in Year 10, I haven't learnt linear algebra. I don't know what ℝ2 or arctan is either. I don't think I need the formulas, right know I just want to understand the concept. What I've got so far is that to represent direction of a vector in 2D You either use angle with the x axis or resolve it along the x and y axes and you can obtain one form from the other. I hope that's correct.

Could someone answer the 2nd question? I need that pretty soon because I want to ask my teacher tomorrow. Last class she drew a s/t graph for uniform circular motion which I didn't think was correct because she just seemed to be taking the magnitude of the vectors and she made all the displacement vectors postive which didn't make sense to me as uniform circular motion was 2D and seemed to need more that just + or -.
 
  • #10
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The 3 unit vectors don't have to be perpendicular to one another (3D space), as long as none of the three can be expressed as a linear combination of the other two. Otherwise, it won't be possible to describe every arbitrary 3D vector as a unique linear combination of the three unit vectors. Use of non-orthogonal unit vectors also usually implies use of a non-orthogonal coordinate system.

You can draw displacement time graphs in 1D, in which case the scalar component of displacement d is the magnitude of the displacement vector d:

d=d ix

where ix is the unit vector in the (positive x-direction) of motion.
 
  • #11
BruceW
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Could someone answer the 2nd question? I need that pretty soon because I want to ask my teacher tomorrow. Last class she drew a s/t graph for uniform circular motion which I didn't think was correct because she just seemed to be taking the magnitude of the vectors and she made all the displacement vectors postive which didn't make sense to me as uniform circular motion was 2D and seemed to need more that just + or -.
for uniform circular motion, the idea is that we keep the distance from the origin constant, right? so then it is only the angle which changes with time.
 
  • #12
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for uniform circular motion, the idea is that we keep the distance from the origin constant, right? so then it is only the angle which changes with time.
She didn't use angular velocity, she just took the initial point and drew displacement vectors from there to points on the circle.
 
  • #13
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The 3 unit vectors don't have to be perpendicular to one another (3D space), as long as none of the three can be expressed as a linear combination of the other two. Otherwise, it won't be possible to describe every arbitrary 3D vector as a unique linear combination of the three unit vectors. Use of non-orthogonal unit vectors also usually implies use of a non-orthogonal coordinate system.

You can draw displacement time graphs in 1D, in which case the scalar component of displacement d is the magnitude of the displacement vector d:

d=d ix

where ix is the unit vector in the (positive x-direction) of motion.
Sorry but didn't get the first para at all. But that's okay, as I go to higher classes I'll learn it, so I won't confuse myself now.:smile:

But as for the second part would that mean that the magnitude would be negative for a vector in the opposite direction to ix? I think I read somewhere the magnitude is always positive or something to that effect.
 
  • #14
BruceW
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the initial point... being the origin? even if she is drawing displacement vectors from the origin to points on the circle, I still think that when she is drawing the s/t graph, the 's' is going to be the angle. (or possibly the constant r times by the angle).
 
  • #15
WannabeNewton
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It has to be the time varying angular coordinate times the constant radius. That's the only degree of freedom there is for such a system; there is nothing else for her to draw as far as the displacement vs. time graph goes.
 
  • #16
Bandersnatch
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She could have drawn the displacement along just one of the cartesian coordinates.
 
  • #17
BruceW
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haha, yeah that is clearly what she would have drawn, since that makes shm, while drawing the angle just gives a straight line. I must be more asleep than I realise :s
 
  • #18
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So I asked her and turns out that what she was plotting was just the magnitude of the displacement on y and time on x.
 

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