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Question Number 6935 by FilupSmith last updated on 05/Aug/16
v=<x_v , y_v >  u=<x_u , y_u >  v and u have angles:  θ_v =tan((y_v /x_v )) and θ_u =tan((y_u /x_u )) respectively  if t=<x_t , y_t > and has angle θ_t =((θ_v +θ_u )/2)  what are x_t  and y_t ? can this be solved?
$$\boldsymbol{{v}}=<{x}_{{v}} ,\:{y}_{{v}} > \\ $$$$\boldsymbol{{u}}=<{x}_{{u}} ,\:{y}_{{u}} > \\ $$$$\boldsymbol{{v}}\:\mathrm{and}\:\boldsymbol{{u}}\:\mathrm{have}\:\mathrm{angles}: \\ $$$$\theta_{{v}} =\mathrm{tan}\left(\frac{{y}_{{v}} }{{x}_{{v}} }\right)\:\mathrm{and}\:\theta_{{u}} =\mathrm{tan}\left(\frac{{y}_{{u}} }{{x}_{{u}} }\right)\:\mathrm{respectively} \\ $$$$\mathrm{if}\:\boldsymbol{{t}}=<{x}_{{t}} ,\:{y}_{{t}} >\:\mathrm{and}\:\mathrm{has}\:\mathrm{angle}\:\theta_{{t}} =\frac{\theta_{{v}} +\theta_{{u}} }{\mathrm{2}} \\ $$$$\mathrm{what}\:\mathrm{are}\:{x}_{{t}} \:\mathrm{and}\:{y}_{{t}} ?\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{solved}? \\ $$
Answered by nburiburu last updated on 04/Aug/16
if the angle of t is an average of  u and v  then you could try  find  cos(θ_t ) and sin(θ_t )  but if you remember adition of vectors  then t could be define as  t =(u/(∥u∥))+(v/(∥v∥))   since both have same length, the diagonal  of the parallelogram with u and v as sides  has the average of angles.    Of course, there is more than one solution  but every solution shall be a proportional  vector to this one.
$${if}\:{the}\:{angle}\:{of}\:{t}\:{is}\:{an}\:{average}\:{of}\:\:{u}\:{and}\:{v} \\ $$$${then}\:{you}\:{could}\:{try}\:\:{find}\:\:{cos}\left(\theta_{{t}} \right)\:{and}\:{sin}\left(\theta_{{t}} \right) \\ $$$${but}\:{if}\:{you}\:{remember}\:{adition}\:{of}\:{vectors} \\ $$$${then}\:{t}\:{could}\:{be}\:{define}\:{as} \\ $$$${t}\:=\frac{{u}}{\parallel{u}\parallel}+\frac{{v}}{\parallel{v}\parallel}\: \\ $$$${since}\:{both}\:{have}\:{same}\:{length},\:{the}\:{diagonal} \\ $$$${of}\:{the}\:{parallelogram}\:{with}\:{u}\:{and}\:{v}\:{as}\:{sides} \\ $$$${has}\:{the}\:{average}\:{of}\:{angles}. \\ $$$$ \\ $$$${Of}\:{course},\:{there}\:{is}\:{more}\:{than}\:{one}\:{solution} \\ $$$${but}\:{every}\:{solution}\:{shall}\:{be}\:{a}\:{proportional} \\ $$$${vector}\:{to}\:{this}\:{one}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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