Vectors A and B are in the xy plane

[DoctorMathU]

Homework Statement

Vectors A and B are in the xy plane and their scalar product is 20.0 units. If Amakes a 27.4° angle with the x-axis and has magnitude A=12.0 units and B has magnitude B= 24.0 units, what can you say about the direction of B?
Answer: 113.4° and 301.4°

The Attempt at a Solution

I am not really good at vectors, so i just did the basic stuff like, cos(x)= a⋅b/(||a||⋅||b||)
But it gives me 86°, that's not the solution...

 

[berkeman]

Homework Statement

Vectors A and B are in the xy plane and their scalar product is 20.0 units. If Amakes a 27.4° angle with the x-axis and has magnitude A=12.0 units and B has magnitude B= 24.0 units, what can you say about the direction of B?
Answer: 113.4° and 301.4°

The Attempt at a Solution

I am not really good at vectors, so i just did the basic stuff like, cos(x)= a⋅b/(||a||⋅||b||)
But it gives me 86°, that's not the solution...

Welcome to the PF.

It usually helps to draw a diagram with the vectors on it to help you set up the calculation. Can you Upload a JPEG copy of your sketch?

 

[DoctorMathU]

Okay, I only know for A.

 

[berkeman]

View attachment 215817 Okay, I only know for A.

What is the Relevant Equation for the scalar product? What does that then tell you about the angle between A and B?

 

[Doc Al]

But it gives me 86°, that's not the solution...

That's the angle between the two vectors.

 

[DoctorMathU]

The relevant equation for the angle between A and B is cos(x)= a⋅b devided by ||a|| *||b||

 

[berkeman]

The relevant equation for the angle between A and B is cos(x)= a⋅b devided by ||a|| *||b||

Did you understand Doc Al's comment?

That's the angle between the two vectors.

You should be able to update your diagram with the possible positions for the B vector now... Please upload your updated diagram. Thanks.