Proof of sin(θ-Φ)=sinθcosΦ-cosθsinΦ using vector algebra

[Jamiemma1995]

Homework Statement

given two unit vectors a= cosθi + sinθi b=cosΦi+sinΦj prove that sin(θ-Φ)=sinθcosΦ-cosΦsinθ using vector algebra[/B]

Homework Equations

sin(θ-Φ)=sinθcosΦ-cosΦsinθ[/B]

The Attempt at a Solution

axb= (cosθsinΦ-cosΦsinθ)k and I'm guessing that the change in sign has something to do with the fact that k is perpendicular to the vectors I'm usingΦ-θ[/B]

 

[haruspex]

You could try a suitable rotation of axes, so that the cross product should still look the same.

 

[ehild]

The Attempt at a Solution

axb= (cosθsinΦ-cosΦsinθ)k and I'm guessing that the change in sign has something to do with the fact that k is perpendicular to the vectors I'm usingΦ-θ[/B]

The vectors a and b have zero "z" component. Remember how the components of the cross product are calculated.

 

[Jamiemma1995]

The vectors a and b have zero "z" component. Remember how the components of the cross product are calculated.

when I calculated the components I got axb = ( cosθi +sinθj )x(cosΦi + sinΦj) axb=cosθcosΦixi + cosθsinΦixj +sinθcosΦjxi +sinθsinΦjxj ixi=1x1xsino=0 jxjxsin0=0 ixj=1x1sin90=1 and jxi=-1 because AxB=-BxA and was then left with axb= cosθsinΦ(1) + sinθcosΦ(-1)=cosθsinΦ -sinθcosΦ but its supposed to be the other way around, I don't understand where I'm going wrong :(

 

[ehild]

was then left with axb= cosθsinΦ(1) + sinθcosΦ(-1)=cosθsinΦ -sinθcosΦ but its supposed to be the other way around, I don't understand where I'm going wrong :(

Think of the other definition of the cross product, how to get its direction applying right-had rule, so sin(θ-φ)=bxa.
https://www.mathsisfun.com/algebra/vectors-cross-product.html

 

[Jamiemma1995]

Think of the other definition of the cross product, how to get its direction applying right-had rule, so sin(θ-φ)=bxa.
https://www.mathsisfun.com/algebra/vectors-cross-product.html

oh I think I get it now so what your saying is although I could find axb and get the answer I got first , I could just as easily choose bxa as they're both the same but have opposite signs and so it still satisfies the proof.

 

[ehild]

oh I think I get it now so what your saying is although I could find axb and get the answer I got first , I could just as easily choose bxa as they're both the same but have opposite signs and so it still satisfies the proof.

The correct one is bxa.

 

[J Hann]

Isn't (i + j) X (i + j) = j X i + i X j)
What is the sign of j X i ?

 

[ehild]

Isn't (i + j) X (i + j) = j X i + i X j)
What is the sign of j X i ?

minus. Why do you ask? It was correct in Post#4.