# 3D vector

1. Jul 8, 2011

### athrun200

1. The problem statement, all variables and given/known data

2. Relevant equations

A=i+j-2k
C=j-5k

3. The attempt at a solution

How to obtain the correct answer?

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2. Jul 9, 2011

### Staff: Mentor

Velocity is a vector tangential at point C to the circular path traced out.

3. Jul 9, 2011

### I like Serena

Hi athrun200!

Let's first get the definition and the relevant formulas in order.

Can you say what kind of rotational movement is made?
That is, what is the axis?

I'm feeling generous, so I'll give you the formula that relates vector speed and angular velocity.

It's:
$$\vec v = \vec \omega \times \vec r$$
Note that these are all vectors and that the multiplication is the cross product.
Are your familiar with the vector representation of an angular velocity?
And are you familiar with the cross product?

4. Jul 9, 2011

### athrun200

I think vector A is the rotation axis.

I am familiar with cross product but I am not sure how to get r.
Is r means the radius of the ratation?

If it is, then both$$\vec \omega$$ and$$\vec r$$ are scalars. ($$\vec \omega$$ =2 which is given)

If both of them is scalars, then the velocity should be scalar too.
However the answer is a vector.

This is what I am not understand.

5. Jul 9, 2011

### I like Serena

Yes, the vector A is the rotation axis.

You seem very eager to turn vectors into scalars. Don't! They're not.
They were marked with an arrow on top for a purpose. They are vectors and not scalars.

The question becomes how to find these vectors.

From your answer I deduce that you are not aware of the vector representation of the angular velocity.
(Since you didn't answer my question and went on converting vectors into scalars.)

So here's a picture from wikipedia that more or less shows what's intended.

However, in this picture r is shown to be a vector starting in the center of the circular motion.
This is not required. r can be a vector that starts on any point of the rotational axis.
I just couldn't find a better picture yet.

The vector of angular velocity is the vector that is parallel with the axis of rotation, and that has a length that is equal to the scalar angular velocity.

Can you write down the vector representation of angular velocity?
And can you say which vector r is intended now?

6. Jul 9, 2011

### athrun200

Oh, it is my first time to know that angular velocity is along the axis of rotation.(When I studied A-Level physics, it didn't mention anything about vector all of them is scalar.)

So, in order to make the angular velocity lies on the axis of rotation, we get 2$\vec{A}$. However, the magnitude is not 2. To ensure the vector has magnitude of 2, we need to have the unit vector of $\vec{A}$.
Thus $\vec{\omega}=2\frac{\vec{A}}{\vec{|A|}}$

I am not sure for $\vec{r}$.
Is $\vec{r}=\vec{C}-\vec{A}$?

7. Jul 9, 2011

### I like Serena

Good! And I see you found the [ itex ] tag.

Vector A is a vector of arbitrary length defining the axis.
Vector C is a vector from the axis to the point you're interesting for finding the velocity vector.
You'll find your result comes out the same if you add any multiple of A to it (since effectively its starting point will still be on the axis)!
To be clear if you take $\vec{r}=\vec{C}$ that will be just fine.

Last edited: Jul 9, 2011
8. Jul 9, 2011

### athrun200

But I still don't understand why I can use both $\vec{r}=\vec{A}+\vec{C}$ and $\vec{r}=\vec{A}-\vec{C}$.

Thx so much

9. Jul 9, 2011

### I like Serena

Not really. I don't have any books myself about it, nor do I know of a website :)
You can try e.g. wikipedia, but perhaps that is a bit too much for you.

Basically it's a property of the cross product, which looks at the part where 2 vectors are perpendicular to each other. Any parallel part is effectively ignored.

You can also see it algebraically. That is:
$$(\vec A + \vec C) \times \vec A = (\vec A \times \vec A) + (\vec C \times \vec A) = \vec 0 + (\vec C \times \vec A)$$
Note that the cross product of a vector with itself is zero.

10. Jul 9, 2011

### athrun200

I understand now, thanks so much.