# Homework Help: Calculate absolute value of A+B

1. Jan 29, 2007

### tnhoots

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

Consider two vectors by A=5i-3j and B=-i - 2j
Calculate A+B
Calculate A-B
Calculate absolute value of A+B
Calculate absolute value of A-B

2. Relevant equations
R= (Ax + Bx)i + (Ay + By)j
R= (Ax + Bx)i - (Ay + By)j

3. The attempt at a solution

I'm not sure where to start with this problem. I know the equations to use. But its figuring out what value goes where that is hindering me

2. Jan 29, 2007

### Kurdt

Staff Emeritus
For the first two you have written the equations you need almost (The subtraction equation needs modifying). What is stopping you plugging in Ax and Bx etc. For parts 3 and 4 do you know how to calculate the absolute value of a vector?

Last edited: Jan 29, 2007
3. Jan 29, 2007

### tnhoots

still working

I found the solution to part A. For part B I am putting the numbers into the equation B+(-A). It looks like this (-1-5)+(-2+3) and I get -6i+1j. However, that appears to be wrong. I don't know how to do the absolute value of a vector. And I also don't know how to find the direction of A+B and A-B.

4. Jan 29, 2007

### Kurdt

Staff Emeritus
When adding vectors we have as you have above:

$$\mathbf{C}=(A_x +B_x) \mathbf{\hat{i}} + (A_y +B_y)\mathbf{\hat{j}}$$

For vector subtraction or vector difference we have:

$$\mathbf{C}=(A_x -B_x) \mathbf{\hat{i}} + (A_y -B_y)\mathbf{\hat{j}}$$

The absolute value is given by pythagorean theorem.

$$|\mathbf{A}| = \sqrt{A_x^2+A_y^2}$$

Last edited: Jan 29, 2007
5. Jan 29, 2007

### tnhoots

Ok...still a little confused on the absolute value thing.
For the absolute value of A+B I found the absolute value of A which is square root of 34=5.83. Then found the absolute value of B which is square root of 5=2.236. Then I added those together. I did the same for the absolute value of A-B except i subtracted them instead of adding them. However, this isn't right. Where am I going wrong?

I am also confused on how to find the direction of A+B and A-B. I thought that you found the direction of something by taking the tangent(-1) of y/x. However, I did that on a problem where y=35.0 and x=-22.5 and got the answer to be -57.3 and this was wrong...Isn't that the correct way though?

6. Jan 29, 2007

### Kurdt

Staff Emeritus
With regards to the absolute value, you have done it wrong. What you need to do is take the absolute value of the new vector which is A+B or A-B.

I'm not sure about your second question because A+B and A-B are vectors they already have direction. Are you talking about finding a specific angle relative to one of the axes?

7. Jan 29, 2007

### tnhoots

So I would take the absolute value of A+B which was 4i+-5j? I would do that by taking the square root of (4squared + -5squared)? And do the same for the absolute value of A-B which was 6i + -1j? Would I subtract those square roots? And the question about their direction wanted to know what there direction would be in degrees from the positive x axis.

8. Jan 29, 2007

### Kurdt

Staff Emeritus
That is correct except you would not subtract the square roots for the A-B vector you would add them as normal.

If you want the angle from the positive x-axis, then taking the inverse tangent of y/x is a method to follow but you have to be careful because you have to add some degrees on depending which quadrant the point is in.

9. Jan 29, 2007

### tnhoots

could you elaborate on finding the degrees from the positive x axis. I am not told what quadrant anything is in...

10. Jan 29, 2007

### tnhoots

I have a similar problem about finding the position from the x-axis. In this problem y is 35.0 and x is -22.5. So wouldn't I take the inverse tangent of 35.0/-22.5? This is my thinking but my answer of -57.3 is incorrect.

11. Jan 29, 2007

### tnhoots

Alright, so I figured out the direction of the A+B and A-B. However, when i do the exact same thing for the second problem where x=-22.5 and y=35.0 it is wrong. Would i not take the inverse tangent of 35.0/-22.5. I find this to be
-57.26 degrees from the positive x-axis. Why isn't that correct?

12. Jan 29, 2007

### HallsofIvy

Are you required to do that? You stated the problem as
"Consider two vectors by A=5i-3j and B=-i - 2j
Calculate A+B
Calculate A-B
Calculate absolute value of A+B
Calculate absolute value of A-B"

None of those require finding a direction.

However, to answer your question, look at the specific x and y values. x= -22.5 and y= 35.0 so you are in the second quadrant. You -57.3 degrees is from the negative x-axis. The angle from the positive x-axis is 180- 57.3= 122.7 degrees.

Last edited by a moderator: Jan 29, 2007
13. Jan 29, 2007

### tnhoots

Yes, I was. However I found the correct solution to that one. However, I have a similar problem that wants to know the degrees from the positive x-axis of a vector with components x=-22.5 and y=35.0. I did the inverse tangent of 35.0/-22.5 and find it to be -57.26 degrees from the positive x-axis. Why isn't that correct?

14. Jan 29, 2007

### Kurdt

Staff Emeritus
Thats explained by HallsofIvy above. The angle you calculated was from the negative x-axis not the positive so you need to correct it.

15. Jan 29, 2007

### tnhoots

THANKS!! I think I may finally be getting the hang of this!