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

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To calculate the magnitude of vector C in the equation C = B - A, the components of vectors A and B must be determined correctly. Vector A has a magnitude of 16 units at an angle of 42 degrees from the y-axis, while vector B has a magnitude of 7 units at 31 degrees from the x-axis. Initial calculations for the components of vector B were incorrect due to the misinterpretation of the angle reference. After correcting the components, the magnitude of vector C was found to be 15.45 units. It is emphasized that vector subtraction does not involve the cross product.
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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
 
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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 ...
 
Oops...

I fixed that. I am getting the correct answer now.
 
Vector subtraction has nothing to do with the cross product.
 

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