Solving Displacement Vectors A & B: Find A+B

[chocolatelover]

Homework Statement

Vector A is 3 m and vector b is 3m. The angle formed by the positive x-axis is 30°. Find A +B

Homework Equations

The Attempt at a Solution

Does this look correct?

i=3cos30
j=3sin30

Vector A+B=(3cos30 +3cos30)+(3sin30+3sin30)
=8.20

magnitude=square root((3cos30 +3cos30)^2+(3sin30+3sin30)^2)
=square root(27+9)=6

theta=tan-1(3sin30+3sin30)/(3cos30+3cos30)=77.55

and if it were vector A-B, the only difference would be that you would take (3cos30+3cos30)+(-3sin30-3sin30) and the magnitude would be square root of (3cos30+3cos30^2)+(-3sin30-3sin30)^2 right?

Thank you very much

 

[kamerling]

The angle formed by the positive x-axis is 30°.

I have no Idea what you mean by this.

 

[chocolatelover]

Sorry.The angle formed by the two vectors on the positive x-axis is 30°. Can you tell me if I did this correctly? Would the magnitude of vector A+B be the square root of (3cos30+3cos30)^2+(3sin30+3sin30)^2
=6
angle=3sin30+3sin30/3cos30+3cos30=179.4

Would the magnitude of A-B be:
square root of (3cos30+3cos30)^2+(3sin30+3sin30)=
6

angle=179.4

But they shouldn't be the same, right? Do you see where I went wrong?

Thank you

 

[kamerling]

That still doesn't tell me where A and B point.
The magnitude of A+B can only be 6 if A and B point in the same direction however, and then of course A+B would have the same direction as A And B and then A-B would be the 0 vector, so your answer can't be right.
The easiest way to do this is to compute the components of A and B in the X and Y directions, then add them, and then convert back to a magnitude and direction.

 

[chocolatelover]

Thank you very much

Regards