# Deriving law of sines from cross product

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• Mr Davis 97

#### Mr Davis 97

I am trying to derive the law of signs from the cross product.

First, we have three vectors ##\vec{A} ~\vec{B} ~\vec{C}## such that ##\vec{A} + \vec{B} + \vec{C} = 0##. This creates a triangle. Then, we label the angles opposite the respective sides as a, b, and c. I am not sure where to go from here... We could take the cross product of each combination of ##\vec{A}## and ##\vec{B}##, but these cross products aren't necessarily equal, so can't set them equal to derive the law of sines... Any help would be appreciated.

I am trying to derive the law of signs from the cross product.

First, we have three vectors ##\vec{A} ~\vec{B} ~\vec{C}## such that ##\vec{A} + \vec{B} + \vec{C} = 0##. This creates a triangle. Then, we label the angles opposite the respective sides as a, b, and c. I am not sure where to go from here... We could take the cross product of each combination of ##\vec{A}## and ##\vec{B}##, but these cross products aren't necessarily equal, so can't set them equal to derive the law of sines... Any help would be appreciated.
If you take ## C \times (A+B+C)=0 ## and ## C \times C=0 ## Then ## |A||C|sin(\theta_1)=|B||C|sin(\theta_2) ## (neglecting a minus sign which is of little significance).Result is ## sin(\theta_1)/|B|=sin(\theta_2)/|A| ##

How do I get ##sin(\theta_3)/|C|## in there? That was my main problem.

How do I get ##sin(\theta_3)/|C|## in there? That was my main problem.
That will show up if you take ## B \times (A+B+C)=0 ## (or ## A \times (A+B+C)=0 ##). You can only do two angles at a time by this method. Comparing two of the equations will tie the 3rd one in there. e.g. ## B \times ## gives you ## sin(\theta_2)/|A|=sin(\theta_3)/|C| ##.

Awesome. Thanks! While the way you put it is very understandable, but I'm not really sure how you came to the conclusion that the cross product of one of the vectors with the sum of the other three would lead you to the proof... Basically, "how would I have thought of that?"

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