How Do You Calculate the Magnitude of Vector C in Vector Subtraction?

[x31fighter]

Homework Statement

C = B - A
A = 16 Units
A-theta = 42 degrees from the y-axis
Ai is pointing in second quadrant

B = 7 units
B-theta = 31 degrees from the x-axis
B is pointing in the third quadrant

Homework Equations

The Attempt at a Solution

I am suppose to find the magnitude of C from the components given. When I attempted it, I did it using trigonometric. That got marked wrong. Just a random question, am I able to do it using cross-product or something like that?

Va-x = -16*sin(42)
va-y = 16*cos(42)

Vb-x = - 7.0*sin(31)
Vb-y = - 7.0cos(31)

Did Pythagorean theorem and combined the x and y and got the following result:
C =15.45

 

[rude man]

C is also a vector, so you need to come up with an angle as well as the correct magnitude.

Va_x and Va_y are correct.
Vb_x and Vb_y are incorrect. 31 deg. is from the x, not the y, axis ...

 

[x31fighter]

Oops...

I fixed that. I am getting the correct answer now.

 

[SteamKing]

Vector subtraction has nothing to do with the cross product.

 

[Nickie]

units
C-theta = 20.23 from the x-axis


I would like to clarify that vector subtraction can only be done by subtracting the components of one vector from the corresponding components of the other vector. The cross product is used for vector multiplication, not subtraction. Therefore, the approach used by the student is correct. However, it is important to note that the components of vector A should be positive, as it is pointing in the second quadrant. Using the given information, the magnitude of vector C can be calculated as follows:

C-x = B*cos(B-theta) - A*cos(A-theta)
C-y = B*sin(B-theta) - A*sin(A-theta)

C = √(C-x^2 + C-y^2)

Substituting the values given in the problem, we get:

C-x = -7*cos(31) - 16*cos(42) = -13.78
C-y = -7*sin(31) - 16*sin(42) = -16.47

Therefore, C = √((-13.78)^2 + (-16.47)^2) = 21.02 units

Hence, the magnitude of vector C is 21.02 units. It is important to note that the direction of vector C can be found by calculating the angle C-theta from the x-axis using the following equation:

C-theta = tan^-1(C-y/C-x) = tan^-1(16.47/13.78) = 49.64 degrees

Since the x-component of C is negative, the angle C-theta will be in the third quadrant. Therefore, the final answer is:

C = 21.02 units
C-theta = 49.64 degrees from the x-axis, pointing in the third quadrant.