# Simple vector question

1. Mar 4, 2006

### hellraiser

What is the difference between the component and resolved parts of a vector? To me both seem the same. Can anyone please explain me with a simple example. Thanx.

2. Mar 4, 2006

### vaishakh

I don't think they have any difference.

3. Mar 4, 2006

### 0rthodontist

Did you read anything that implied there was a difference?

4. Mar 5, 2006

### hellraiser

I was doing a problem that seemed to make me believe there was a difference. The definition they have given is:
Given a diagonal we can draw infinite number of parallelograms. Each pair of sides will give a pair of components.
If we are given a vector and we find component of vector in given directions such that they are equivalent to the given vector, this is resoultion of vectors.

The problem I was doing was
Two vectors P and Q have resultant R. The resolved part of R in direction of P is Q. If A be the angle between the vectors prove that
sin (A/2)=sqrt(P/2Q)
I got the answer. But I still don't understand why I took the resolved part of R in direction of P to be R cos(x) .. x is the angle which R makes with P.

5. Mar 5, 2006

### vaishakh

Normally if the angle between two vector is A, then the component of one in the direction of other is cosA times its magnitude. You can compare this with resolution along two axial planes. Let us consider a vector of magnitude x and making angle y with X-axis. Then it makes 90-y with Y-axis.
Thus vector = xcosyi^ + xcos(90-y)j^ = xcosyi^ + xsinyj^

6. Mar 5, 2006

### vaishakh

That is true. You can resolve a vector in to components in infinite number of ways
Here directions are given and hence there is only one way. But if we are the ones who will be selecting the directions, then again there wil be infinite resolutions.

Normally if the angle between two vector is A, then the component of one in the direction of other is cosA times its magnitude. You can compare this with resolution along two axial planes. Let us consider a vector of magnitude x and making angle y with X-axis. Then it makes 90-y with Y-axis.
Thus vector = xcosyi^ + xcos(90-y)j^ = xcosyi^ + xsinyj^

Sorry for the mistake. This would be convenient to read.

7. Mar 6, 2006

### hellraiser

Then the textbook has also given this formula (with derivation) :)

P = R sin(y)/sin(a)
y is the angle between R and other vector Q and a is the angle between P and Q. So what I have come to know after solving problems is that if we are given both the directions along which we have to find components then use this formula or else use the cos times magnitude.
Am I right?