Vector Subtract Given Magnitude

[Lori]

Homework Statement

Let's say i was given Vector A and B. The Angle between them is 60 degrees. Vector A's magnitude is 40 and Vector B's magnitude is 50. Find magnitude of vector C, if C = vector A - vector B.

Homework Equations

I'm given the magnitudes, and need to find magnitude of C

The Attempt at a Solution

I'm kinda confused cause I don't think this is a simple problem. But, I thought that magnitude of C would be 10. Since A-B = -10 , and 10 is the magnitude[/B]

 

[George Jones]

Can you draw a diagram for the situation?

 

[Lori]

 

[Delta2]

By which side of the triangle the vector A-B is represented? What do you get if you properly use law of cosines?

 

[RedDelicious]

Magnitudes don't add that way unless the vectors are parallel and so you need to solve the triangle using the law of cosines.

For example, if the angle between A and B were 0, then by the law of cosines

$$c^2=a^2+b^2-2ab\cos{(0)}= (a-b)^2 \\ c = |a-b|$$

Which is what you tried to do but the angle is not zero in this case which is why it's wrong

 

[Lori]

Magnitudes don't add that way unless the vectors are parallel and so you need to solve the triangle using the law of cosines.

For example, if the angle between A and B were 0, then by the law of cosines

$$c^2=a^2+b^2-2ab\cos{(0)}= (a-b)^2 \\ c = |a-b|$$

Which is what you tried to do but the angle is not zero in this case which is why it's wrong

Thanks. I definitely did not learn this from my physics class.

 

[YarnMonkey]

I thought of this problem as C = A + (-B). This means that it is the same as adding a vector, just the vector you are adding is negative. In order to add the vectors, we must split them into their components. We know that vector A is at an angle of 60 and therefore has X and Y components, while B is parallel to the horizontal and only has an X component equal to its magnitude.

To split vector A, we use Cos and Sin to find the vertical and horizontal components of the vector.
X Component (horizontal): Cos(60)*40=20
Y Component (vertical): Sin(60)*40=34.64

Now that we know the vertical and horizontal components of vector A, we can add the corresponding components of B, although since we are ultimately subtracting B from A, we just make B negative.

This leaves us with:
X: 20 + (-50) = -30
Y: 34.64 + 0 = 34.64

Then we use the Pythagorean theorem to find the result vector of these new horizontal and vertical components
(A*A) + (B*B) = (C*C) = √(-30*-30) + (34.64*34.64) = C
C = 45.83

We know that the resultant vector has a magnitude of 45.83.