Help with A+B and 2A+3B: Magnitude & Direction

[rsixtyone]

I'm don't know how to do this at all, can anyone help me please?

If $$A=[A\angle\theta_{A}]$$ and $$B=[B\angle\theta_{B}]$$,

1) what is the magnitude of A+B?
2) what is the direction of A+B?
3) what is the magnitude of 2A+3B?
4) what is the direction of 2A-3B?

 

[Doc Al]

Start by finding the x and y components of each vector, then operate on those components to find the resultants. You'll need to know that $A_x = A cos\theta_A$ and $A_y = A sin\theta_A$.

 

[marlon]

also keep in mind that once you know the components (x,y,z) of a vector, you can easily calculate the vector's magnitude using the formula sqrt(x²+y²+z²)

just as an addendum to Doc Al's words

regards
marlon

 

[rsixtyone]

This is a new to me, I still couldn't get it. Solve and explain the problem to me Doc? Thank you.

 

[Doc Al]

Poke around here: http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec2

I'll do the first one:
$\vec{R} = \vec{A} + \vec{B}$
First the x components:
$R_x = A_x + B_x = A cos\theta_A + B cos\theta_B$
Then the y components:
$R_y = A_y + B_y = A sin\theta_A + B sin\theta_B$

Thus the magnitude of A + B = $\sqrt{(R_x^2 + R_y^2)}$