# Concerning Vectors in Scalar Form

• Cosmophile
In summary, the conversation discusses the use of vectors in problem solving and how they can be written in scalar form using only magnitude without considering direction. This can cause confusion, but the use of polar coordinates and unit vectors can help clarify the relationship between vector magnitudes and components in different directions.
Cosmophile
Hey, all. I have a question concerning the treatment and use of vectors when solving problems (or in general, really).

I know that vectors have both magnitude and direction, while scalars only have magnitude. However, in solving problems and looking at how others have solved them, I've noticed something: vectors seem to be often written in "scalar form," if you will. For example, consider circular motion

$$\vec r = r(\cos \omega t \hat{\imath}+ \sin \omega t \hat{\jmath})$$
$$| \vec r| = r = constant$$
$$\vec v = \frac {d \vec r}{dt} = r \omega (-\sin \omega t \hat{\imath} + \cos \omega t \hat{\jmath})$$

This all makes total sense to me. Where I find trouble, however, is that when I look around, I see people solve problems by simply saying ##v = r \omega##. No vector notation, no mention of direction, etc. How do they know to do this, and how can I know when to do this as well? This is causing me an unbelievable amount of frustration. Thanks!

They're working with the magnitude of the vectors i.e. ## \vec S=S_x \hat i+S_y \hat j \Rightarrow S\equiv |\vec S|=\sqrt{\vec S\cdot \vec S}=\sqrt{S_x^2+S_y^2} ## which is a scalar. Sometimes you don't need to care about the direction of the vector and considering its magnitude is enough. I suggest you post one problem where you've seen this in the homework section and ask people why you don't need to consider the direction of the vector. You'll understand things better.

Shyan said:
They're working with the magnitude of the vectors i.e. ## \vec S=S_x \hat i+S_y \hat j \Rightarrow S\equiv |\vec S|=\sqrt{\vec S\cdot \vec S}=\sqrt{S_x^2+S_y^2} ## which is a scalar. Sometimes you don't need to care about the direction of the vector and considering its magnitude is enough. I suggest you post one problem where you've seen this in the homework section and ask people why you don't need to consider the direction of the vector. You'll understand things better.

Right, I could tell they were dealing with the magnitude. I'll find a problem and post it later to figure out why this is the case. Thanks.

In addition to what Shyan had to say, your initial equation can be expressed in polar coordinates using unit vectors in the r and theta directions, ##\hat{r}## and ##\hat{\theta}##.

Using ##r=\sqrt{x^2+y^2}## and ##\tan{\theta}=x / y##, a generic vector in polar coordinates is expressed as ##\vec{r}=r \hat{r} + r \theta \hat{\theta}##.

In your problem, ##\vec{r}=r \hat{r} + r \omega t \hat{\theta}##, where ##\omega = \frac{d\theta}{dt}##.

If both ##r## and ##\omega## remain constant in time, ##\vec{v} = 0\hat{r} + r\omega \hat{\theta}## or ##\vec{v} = r\omega \hat{\theta}##.

##\vec{v}## and ##\hat{\theta}## point in the same direction so that ##v \hat{\theta}= r\omega \hat{\theta}##.

We can divide through by the unit vector ##\hat{\theta}##, so that ##v=r\omega## but in the process lose directional information.

The rather subtle point I am trying to relate, is that ##v=r\omega## is more than just an equation in vector magnitudes, it is also an equation between the elements of two vectors.

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Cosmophile
stedwards said:
In addition to what Shyan had to say, your initial equation can be expressed in polar coordinates using unit vectors in the r and theta directions, ##\hat{r}## and ##\hat{\theta}##.

Using ##r=\sqrt{x^2+y^2}## and ##\tan{\theta}=x / y##, a generic vector in polar coordinates is expressed as ##\vec{r}=r \hat{r} + r \theta \hat{\theta}##.

In your problem, ##\vec{r}=r \hat{r} + r \omega t \hat{\theta}##, where ##\omega = \frac{d\theta}{dt}##.

If both ##r## and ##\omega## remain constant in time, ##\vec{v} = 0\hat{r} + r\omega \hat{\theta}## or ##\vec{v} = r\omega \hat{\theta}##.

##\vec{v}## and ##\hat{\theta}## point in the same direction so that ##v \hat{\theta}= r\omega \hat{\theta}##.

We can divide through by the unit vector ##\hat{\theta}##, so that ##v=r\omega## but in the process lose directional information.

The rather subtle point I am trying to relate, is that ##v=r\omega## is more than just an equation in vector magnitudes, it is also an equation between the elements of two vectors.

Could you elaborate on what you mean by the "elements" of the vectors? Perhaps that's just terminology I'm not familiar with.

I'm sorry. Element is very nonstandard usage. I find only two textual references: one to the components of a tensor and another to the components of a symbol. The former is probably in error, ;p.

I should have said "components of a vector". For ##V = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}##, I refer to ##V_x##, ##V_y##, and ##V_z##.

No worries! I figured that was what you meant, based on the equations. I just wanted to be sure.

Out of curiosity, why do we do this if we do lose the directional information?

No, problem. Thank's for asking. I'll wash my mouth out with proper-physics-terminology soap.

Given just ##v=r\omega##, all we can infer is that v is a speed, not really a velocity in the ##\phi## direction. It could mean ##v\hat{r}=r\omega \hat{r}##, for all we know. We only know these are components of a vector pointing in the ##\phi## direction because of the problem set-up.

Also, it could be for clockwise rather than counter clockwise rotation: ##\hat{v (-\phi)} = r\omega \hat{(-\phi)}##.

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## 1. What is a vector in scalar form?

A vector in scalar form is a mathematical representation of a vector using only scalar quantities, such as magnitude and direction. It is often denoted as a column matrix with the magnitude of the vector in the first row and the direction in the second row.

## 2. How is a vector in scalar form different from a vector in component form?

A vector in scalar form only uses scalar quantities, while a vector in component form includes both magnitude and direction components. Scalar form is often used for simplification and convenience, while component form is used for more precise calculations and analysis.

## 3. How do you convert a vector in component form to scalar form?

To convert a vector in component form to scalar form, you can use the Pythagorean theorem to find the magnitude of the vector and trigonometric functions to find the direction. The magnitude will be the first row of the scalar form, and the direction will be the second row.

## 4. What are some real-world applications of vectors in scalar form?

Vectors in scalar form are commonly used in physics, engineering, and computer graphics. They can be used to represent forces, velocities, and other physical quantities. They are also used in computer graphics to represent the position and orientation of objects.

## 5. Can a vector in scalar form have negative values?

Yes, a vector in scalar form can have negative values. The magnitude of the vector can be negative, indicating a direction opposite to the positive direction. This is often represented by a negative sign in front of the magnitude in the first row of the scalar form.

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