Calculate a vector wich forma an angle alpha with another vector

AI Thread Summary
To calculate a vector that forms an angle α with a given 2D vector v, one can use the relationship involving the dot product, where v_x * u_x + v_y * u_y = ||u|| * cos(α). By defining the vector u as u_x * i + u_y * j and considering it as a unit vector (where v_x² + v_y² = 1), the equations can be simplified. The solution requires solving these equations for the components of u. The discussion highlights the importance of correctly applying the equations to find the desired vector. Understanding these relationships is crucial for successfully determining the vector that meets the angle requirement.
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Homework Statement



Looks very simple, but I need help.
I have a 2D vector, v, and I need to calculate a vector which forms an angle \alpha with it.

Thank you very much

Homework Equations



I think that the equations should be, for 2 given vector u and v:
\dpi{150}%20\vec{v}%20\cdot%20\vec{u}%20=%20|\vec{v}|%20\cdot%20|\vec{u}|%20\cdot%20cos%20\alpha.png


png.latex?\dpi{150}%20\vec{v}%20\cdot%20\vec{u}%20=%20v_x%20\cdot%20u_x%20+%20v_y%20\cdot%20u_y.png


thanks
 
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There are, of course, an infinite number of such vectors- of different lengths as well as on either side of u. If you are given vector \vec{u}= u_x\vec{i}+ u_y\vec{j}, taking \vec{v}= v_x\vec{i}+ v_y\vec{v}, then we must have, as you say, v_xu_x+ v_yu_v= \sqrt{u_x^2+ u_y^2}cos(\alpha).

There is no "\sqrt{v_x^2+ v_y^2}" in that because I have decided, for simplicity, to look for a unit vector making angle \alpha with \vec{u}. And, of course, that means that v_x^2+ v_y^2= 1.

Solve those two equations for u_x and u_y.
 
thanks for the help, HallsofIvy, now I see my error. I wasn't using the second equation
 
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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