# Homework Help: Physics Questions About Vector

1. Oct 3, 2004

### soul814

Vector A is 3.00 units in length and points along the positive x-axis. Vector B is 4.00 units in length and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the vectors.
Find
A) A+B
B) A-B

I don't get it.

2. Oct 3, 2004

### arildno

Welcome to PF!
Post in some more detail why you don't get it.

3. Oct 3, 2004

### soul814

Well I'm taking ap physics right now and I dont get it my teacher just puts two problems on the board every day and then goes on solving them. He doesnt teach, so far I dont get any thing in physics.

Thats a question from my textbook. Is A) 3+-4? That seems wrong. Since R = Vector of (A) + Vector of (B)

but isnt the vector of A = 3?
since 3cos90 = 3

How would you solve this problem?

4. Oct 3, 2004

### Pyrrhus

Use the paralelogram addition of vectors.

5. Oct 3, 2004

### soul814

I have no clue what to do? I read the chapter three times and I just dont get it. I know what hte parallelogram addition of vectors is.

Its when you make a diagonal line between the two constants.

Can someone solve it with an explanation for me? Since even if you say all these rules I still wont get it. I have this text book

6. Oct 3, 2004

### arildno

All right, let's take it easy!
1. Suppose you draw an arrow on a piece of paper.
That arrow has two features:
a) It has some length
b) It has a direction
You get that?

7. Oct 3, 2004

### soul814

yep, the coordinates of X would be (0,0) to (0,3)
the coordinates of Y would be (0,0) to (0,-4)

8. Oct 3, 2004

### arildno

Good!
So the arrowhead of X lies at (0,3) and the arrowhead of Y at (0,-4)
Do you agree with this?

9. Oct 3, 2004

### soul814

yes i got that much

10. Oct 3, 2004

### geometer

A useful property of vectors is that they remain the same vector if you move them as long as you don't change their magnitude or direction. So, you can move vector B out along the positive x axis until its "foot" is at the "head" of vector A, and it's still the same vector. Now, what would the vector be that you could draw from the origin to the tip of vector B in it's new position?

11. Oct 3, 2004

### soul814

(3,0) to (3,-4) so it would look like

--->
....|
....|
....|
....|
....\/

12. Oct 3, 2004

### arildno

Allright!
At this position, both X and Y has the "foot" at the origin.
That means:
"Slide" the foot of Y along X (without rotating Y!), such that when the foot of Y coincides with the arrowhead of X, you've got a line segment parallell to Y attached to the arrowhead of X (that line segment (the "translated Y") is of equal length as Y).
Draw this.

We say that the SUM of X and Y is the vector which has its foot in the origin, and its arrowhead coincident with the arrowhead of "translated Y"

Get that?

13. Oct 3, 2004

### geometer

Looks like I was a day late and a dollar short here! Sorry for jumping in the middle of this!

14. Oct 3, 2004

### soul814

like that right? and then you draw a line? and use the pythagorean theorem

--->
\...|
.\..|
..\.|
...\|
....\/

15. Oct 3, 2004

### arildno

Precisely!
This is the graphical way in summing two vectors together; how can you do this arithmetically if you're given the vector's components?

16. Oct 3, 2004

### soul814

a squared + b squared = c squared?
3*3 + (-4)*(-4) = 25

square root of 25 = 5?

17. Oct 3, 2004

### arildno

The Sum of (3,0) and (0,-4) is given by:
(3,0)+(0,-4)=(3+0,0-4)=(3,-4)
(It seems you did this somewhere..)
The magnitude (length) is, as you said, 5.
What is the vector's direction (given, for example, as the angle the vector makes with the x-axis)

18. Oct 3, 2004

### soul814

45 degrees, going southeast, Oh when they ask what is A + B they are asking for the point at which it ends

19. Oct 3, 2004

### arildno

45 degrees are incorrect; think again.

20. Oct 3, 2004

### soul814

negative 45?

21. Oct 3, 2004

### arildno

No, why did you think it was 45 in the first place?

22. Oct 3, 2004

### soul814

3-4-5 triangle?

23. Oct 3, 2004

### arildno

That's definitely correct, but there's no 45 degrees angle in that triangle.
Lets look at the tangent value in a 3-4-5 angle:
You have:
$$tan(\theta)=\frac{oppositeside}{adjacentside}$$
In your case (let's just look at "4" rather than "-4") you have:
$$tan(\theta)=\frac{4}{3}$$
Agreed?
now, what angle $$\theta$$ satisfies that equation?
(You'll need to use your calculator)

24. Oct 3, 2004