# Homework Help: Projection of vector vs vector components

1. Apr 7, 2014

### norstudent

1. The problem statement, all variables and given/known data
This is an example problem where you have a force F at 100N applied at an angle of 45 degrees from a horizontal u-axis. You have the u-axis at zero degrees, then 45 degrees after that you have the Force then 15 degrees after th at you have the v-axis

You are asked to determine the magnitudes of the projection of the force onto the u and v axes
The solution is provided, and this part of the problem I understand

However, after the solution is found there is a comment that I do not understand, it says:
"NOTE: These projections are not equal to the magnitudes of the components of force F along the u and v axes found from the parallelogram law

I am trying to understand why this is NOT the case
Perhaps I have misunderstood what the projection of a vector is, I thought that when we project a vector along two lines, we find the components of it along those two lines, so shouldn't this equal the components of the vector along those same lines found from the parallelogram law?

If it's any help, here's a screenshot of the example, the part where I am confused is highlighted:
https://www.dropbox.com/s/uc92zr1vyyaurmy/projectionofvector.png

2. Relevant equations
A*B=ABcosθ

3. The attempt at a solution
By decomposing the force of 100 N along the v and u axes respectively, using the parallelogram law I have found the force-component along the u-axis to equal 29.89N and the force-component along the v-axis to be 81.65N

This proves that the projections of the vector onto the two axes u and v, do not equal the projections of the force F along these two axes which are stated in the solution for the exame to be 70.7N for the u-axis and 96.6N for the v axis

So the statement at the bottom is indeed correct

I assume then that I have a faulty understanding of what the projection of a vector is

Paraphrasing the definition of the term "projection of a vector onto a line" the same textbook says that:
The component of the a vector A parallel to a given line aa is equal to the magnitude of the vector times cosine of the angle between the vector and the line

To me this seems just like finding the component of the vector along that line, so if you do it for two lines it would be just like finding the vector components along two axes, like we do when we apply parallel law theory

This is where I am stuck, I know I am understand this wrong somehow, please point what it is that I am understanding wrong?

Thanks!

Last edited: Apr 7, 2014
2. Apr 7, 2014

### az_lender

You, and all of us, are accustomed to situations where the coordinate axes are perpendicular to one another. Where coordinate axes are not perpendicular, the definition of a "projection," which you have correctly stated, simply cannot produce the same result that the parallelogram law produces (unless the vector you are projecting onto an axis is already parallel to that axis). And in fact that is the whole point of this example.

3. Apr 7, 2014

### Staff: Mentor

To add to this, it is worth mentioning that, if you tried to use the perpendicular projections onto the u and v axes to resolve the vector F in terms of components in the u and v directions, you would get the wrong answer. The correct resolution of the vector F into components in the u and v directions is the parallelogram law.

Chet

4. Apr 8, 2014

### norstudent

Thanks for the replies guys

So when doing the vector projection onto the first line, it is like you are setting up a coordinate system at that line with two vertical perpendicular axes, and the projection of the vector onto this line is the vertical component and likewise for the second line, so these two vectors are not in any ways the components of F- to find those we must apply the parallelogram law, whereas if these two lines had an angle of 90 degrees between them, we could use the vector projection as well as parallelogram law?

5. Apr 8, 2014

### Staff: Mentor

Yes.

6. Sep 29, 2014

### srg

Hi guys,

It seems I've run into the same confusion as the OP. In fact, it seems I have the same exact textbook as well. I've read and reread the answers here, as well as the definitions and explanations in my textbook, however I'm still not understanding this properly.

I also used the parallelogram law to find the components of $\vec{F}$ along the u and v axes:
http://srg.sdf.org/images/PF/VecProj.png [Broken]
$\frac{\sin{120}}{100} = \frac{\sin{45}}{v} \therefore v=81.65 N$
and
$\frac{\sin{120}}{100} = \frac{\sin{15}}{u} \therefore u=29.89 N$

As the OP pointed out, this differs from the answer given for the projection of $\vec{F}$. Those answers above are the components of $\vec{F}$, correct?
[1]: The "first line" meaning the u axis?

I apologize for bringing up this old thread. My textbook doesn't do a good job on describing the difference between components and projection.

Thank you!

Last edited by a moderator: May 7, 2017
7. Sep 29, 2014

### Staff: Mentor

I would to this problem differently. Are you familiar with the concept of unit vectors?

Chet

8. Oct 3, 2014

### srg

I am.

I've read and reread this, along with my textbook, yet again, and I'm just not getting it. We're already past this in my class, so I'm just going to forget it and move on.

Thanks.

9. Oct 3, 2014

### Staff: Mentor

Sorry to hear this. If you ever decide to come back to this problem, please try this:

1. Express the unit vector in the u direction in terms of the unit vectors in the x and y directions
2. Express the unit vector in the v direction in terms of the unit vectors in the x and y directions
3. Express the force F in terms of the unit vectors in the x and y directions
4. Solve for the unit vectors in the x and y directions in terms of the unit vectors in the u and v directions
5. Substitute these results into your equation for the force F to obtain F in terms of the unit vectors in the u and v directions.

Chet