Master the Art of Dividing Vectors for Scalar Quotients with These Simple Steps

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Dividing vectors to produce a scalar quotient requires calculating scalar products rather than directly dividing vectors. The correct approach involves computing the dot products of the vectors, yielding scalars that can then be divided. Specifically, one can express this as the ratio of the dot products, such as \(\frac{\vec{u}\cdot \vec{u}}{\vec{v}\cdot\vec{v}}\). Taking the square root of this ratio gives a measure of the lengths of the two vectors. However, it's important to note that vector division is not typically defined and may not yield desirable properties.
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If I want to divide vectors and produce a scalar quotient can I go as follows:
\frac{\vec{u}}{\vec{v}} \cdot \frac{\vec{v}}{\vec{v}}
i.e. compute the dot products and then divide
 
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Well you do not have many options. In order to have a scalar product , you need either

a)two vectors or
b)two numbers or
c)a vector and a number

that you multiply...

The quotient of two vectors is NOT a vector nor a number. The only thing that you can do is first calculate the scalar product in the numerator and then the scalar product in the denominator. This yields two numbers (ie scalars) that you can devide...


marlon
 
Thank you.
 
In other words, \frac{\vec{u}\cdot \vec{u}}{\vec{v}\cdot\vec{v}}.

In fact, I might be inclined to take the square root of that:
\sqrt{\frac{\vec{u}\cdot \vec{u}}{\vec{v}\cdot{\vec{v}}}.
so that you are really dividing the lengths of the two vectors.

Of course, that will not have very nice properties. Division of vectors is not normally defined. What are you doing this for?
 
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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
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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