How do i solve this physics problem that contains vectors

[baseball2233]

Homework Statement

Each of the displacement vectors A and B have a magnitude of 3.00.
Vector B lies on the y-axis and Vector A has a 30 degree angle from the positive X axis.

a) A+B
magnitude = ?
θ = ?

(b) A-B
magnitude = ?
θ= ? °

(c) B - A
magnitude =?
θ= ? °

(d) A - 2B
magnitude =?
θ= ? °

Homework Equations

The Attempt at a Solution

I really don't have a clue how to figure this out, any help would be great

 

[nicksauce]

The idea is, when adding vectors, to separate them into their components, and then add those components to make the new vector. You know $$\vec{B}=3\hat{y}$$, and $$\vec{A}=3\cos{30}\hat{x} + 3\sin{30}\hat{y}$$, at least you should know that. So then, for example, $$A + B = 3\cos{30}\hat{x} + (3 + 3\sin{30})\hat{y}$$. Make sense? See where you can get using this method.

 

[baba89]

I would suggest approaching this problem by breaking down the given information into smaller, more manageable pieces. First, let's draw a diagram to visualize the situation.

We know that Vector B lies on the y-axis, which means it has a 90 degree angle from the positive X axis. Vector A has a magnitude of 3.00 and an angle of 30 degrees from the positive X axis. This can be represented on a graph as shown below:

Now, let's look at the given equations and try to solve for the unknown values.

a) A+B
To find the magnitude of A+B, we can use the Pythagorean theorem: c^2 = a^2 + b^2. In this case, a and b represent the x and y components of the vectors A and B respectively. Using basic trigonometry, we can find the x and y components as follows:

a = Acos(30) = 3cos(30) = 2.598
b = Bsin(90) = 3sin(90) = 3

Plugging these values into the Pythagorean theorem, we get:
c^2 = (2.598)^2 + 3^2 = 6.744
c = √6.744 = 2.596

Therefore, the magnitude of A+B is 2.596. To find the angle θ, we can use the inverse tangent function (tan^-1) on the ratio of the y and x components:

θ = tan^-1(b/a) = tan^-1(3/2.598) = 51.34°

b) A-B
Using the same method as above, we can find the magnitude and angle of A-B as follows:

a = Acos(30) = 3cos(30) = 2.598
b = -Bsin(90) = -3sin(90) = -3

Plugging these values into the Pythagorean theorem, we get:
c^2 = (2.598)^2 + (-3)^2 = 6.744
c = √6.744 = 2.596

Therefore, the magnitude of A-B is also 2.596. To find the angle θ, we can use the inverse tangent function (tan^-1) on the ratio of the y and x components:

θ = tan^-1