# Homework Help: Please need help

1. Sep 26, 2007

### kevinf

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

in the sum vector a + vector b = vector c, vector a has a magnitude of 12.0m and is angled 40.0 degrees counterclockwise from the +x direction, and vector c has a magnitude of 15.0m and is angled 20.0 degrees counterclockwise from the -x direction

a)what is the magnitude
b) what is the angle (relative to +x) of vector B.

3. The attempt at a solution

i first found the x and y component of each vector:

so for vector a the x component would be 12 cos(40) and the y component would be 12 sin(40). for vector C the x component would be 15 cos (20) and the y component would be 15 sin (40). so do i just add the x components of the 2 vectors and the y component of the 2 vectors, do pythagorean theorem and find out what the magnitude is.
so would i be (12cos(40)*15cos(20))^2+(12sin(40)+15sin(20))^2 and then square root?

Last edited: Sep 26, 2007
2. Sep 26, 2007

### mgb_phys

That's the right idea.
Draw a triangle, where the first side is 12m at 40deg above X, then from the end of that draw a line 15m at 20deg below X, then draw a closing line back to the start - this is the resulatant vector.

3. Sep 26, 2007

### Avodyne

Your problem said the second vector is at a 20 degree angle counterclockwise from the minus x direction. If that's not a typo, you need to think about whether the components of the second vector are positive or negative.

4. Sep 26, 2007

### kevinf

yeah, its not a typo, so the second vector will have negative x and y right? does the problem change if vector b is the unknown not vector c.
so what do you guys get, i am just checking my answers. i got 26.59 for the magnitude and 28.87 degrees. here's my work just to clarify my explanation

Last edited: Sep 26, 2007
5. Sep 26, 2007

### kevinf

bump...

6. Sep 26, 2007

### Avodyne

If b is unknown, then b = c-a, that is, you subtract the components of a from the components of c. So your magnitude above looks like it is correct.

7. Sep 26, 2007

### kevinf

but i don't think i subtracted it though, i added Ax and Bx and Ay and By. so my answer in the pic is correct? and i think i get the same answer even if i subtracted

Last edited: Sep 26, 2007
8. Sep 27, 2007

### Avodyne

Your Cx and Cy should both be negative, not positive (which should be clear from your picture). Then, since you are subtracting the positive A components from the negative C components, both terms in both Rx and Ry should be negative. Then, your magnitude of R is correct (I didn't actually check the numbers), but your angle is off by 180 degrees, since both Rx and Ry are negative.

9. Sep 27, 2007

### kevinf

so i add 180 to my theta since both rx and ry are negative going to third quadrant. and the R would be the same because i square it later in the pythagorean theorem, making it positive right?

10. Sep 27, 2007

### kevinf

bump....

11. Sep 27, 2007

### Avodyne

Right! (The magnitude of a vector is always positive.)