What is the result of combining vectors A and B?

[queenspublic]

Vector A has a magnitude of 2.00 m and is directed east. Vector B has a magnitude of 6.00 m and is directed 24.0° west of north.

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

 

[NBAJam100]

Vector A has a magnitude of 2.00 m and is directed east. Vector B has a magnitude of 6.00 m and is directed 24.0° west of north.

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

Break the vectors into their <x,y> components and then do the indicated operations.

 

[queenspublic]

How do I break it? Do I use sine and cosine?

 

[queenspublic]

Add two vectors. Find the x and y components of the two vectors. Add the x components to get the net x component. Do the same for the y components. Draw the net x component and the net y component. Magnitude is determined with the Pythagorean theorem, and angle is determined with an inverse tangent. If, instead, you add the vectors on a vector-capable calculator, all angles must be relative to the positive direction of the x axis.

 

[NBAJam100]

How do I break it? Do I use sine and cosine?

Yup. your first vector is already in component form because it is just pointed east, so its of the form <2,0>. (remember sign of the number matters in this).

For your second one if you draw it out you'll notice you have a triangle. Use your sin and cos functions to find the X and Y components of the second vector.

Then you'll have vector A broken into x and y =<x1,y1> and vector B broken into x and y <x2,y2> and then you can do the operations to them. Remember when you are doing the operations you can only add x to x and y to y.

<1,1>+<2,5> = <3,6>

[EDIT] and yes, about the angle with respect to the origin it brought up inverse tangent. That is because tan=(opp/adj)... which is pretty much the same thing in this case as tan(theta)=(y/x) so the angle w.r.t the positive x-axis is theta=arctan(Y/X)