Cross Product of a Constant Vector

AI Thread Summary
The discussion revolves around the cross product of a constant vector and another vector, clarifying that the term "constant" does not affect the cross product's definition. The cross product is calculated using the determinant of a matrix formed by the two vectors, applicable to both constant and variable vectors. There is confusion regarding the initial question, which was previously addressed in another thread. The participant acknowledges a mix-up between the cross product and the curl of a vector, leading to their apology for any frustration caused. Ultimately, the key takeaway is that a constant vector behaves like any other vector in the context of cross products.
quantumfoam
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Okay, now that my question has been cleared up, what is the cross product of a constant vector and a vector? Is there a formula?
 
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To cross two vectors you can simply take the determinate of the matrix they make. As far as I know this method works for all constant and variable vectors which have a determinate that exist.

 
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quantumfoam said:
Okay, now that my question has been cleared up, what is the cross product of a constant vector and a vector? Is there a formula?
What do you mean by "the cross product of a constant vector and a vector"? The cross product is a product of two vectors. Whether "constant" or "variable" has nothing to do with the product.

And what question are you referring to with "now that my question has been cleared up"? This is exactly the question you asked in https://www.physicsforums.com/showthread.php?t=668424. If you did not understand the answers there, explain what you do not understand. Do not just start a new thread for exactly the same question!
 
Sorry if I have made SOME people upset. I wasn't really thinking about the math. Nevermind about this problem. A constant vector could be just like any other vector. I somehow confused the cross product with the curl of a vector. Sorry about my confusion everyone. Once again, I am sorry for making anyone upset and frustrated. Just some confusion HallsofIvy.
 

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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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