Let u be a constant vector and v be a vector of constant magnitude such that |v| = 1/2 |u| and |u| not equal to 0. Then the maximum possible angle between u and u + v is?

Dear Student,

Please find below the solution to the asked query:

Consider the vector v makes an angle θ with the negative side of vector u. So, the resultant component in x direction is,

$$\begin{gathered} {\left(u+v\right)}_x=u-v cos \theta \Rightarrow {\left(u+v\right)}_x= u-\frac{u}{2}cos \theta \\ \Rightarrow {\left(u+v\right)}_x= \frac{u}{2}\left(2-\cos \theta \right) \end{gathered}$$

and the y component of the resultant vector is,

${\left(u+v\right)}_y=v \sin \theta \Rightarrow {\left(u+v\right)}_y= \frac{u}{2}\sin \theta$

According to the parallelogram law of vector addition, the angle between the resultant with the first vector is,

$\tan \alpha =\frac{{\displaystyle \frac{u}{2}} \sin \theta }{{\displaystyle \frac{u}{2}}\left(2+ \cos \theta \right)}=\frac{{\displaystyle sin \theta }}{{\displaystyle \left(2+ cos \theta \right)}}$

If α has to be maximum, when the $\frac{d\alpha }{d\theta }$ should be equals to zero. Therefore,
$$\begin{gathered} \frac{d}{d\theta }\left\{\tan \alpha \right\}=\frac{d}{d\theta }\left\{\frac{\sin \theta }{2+\cos \theta }\right\} \\ \Rightarrow se{c}^2\alpha \left(\frac{d\alpha }{d\theta }\right)=\frac{sin \theta \left\{-\sin \theta \right\}-\left(2+cos \theta \right)\left\{\cos \theta \right\}}{{\left(2+cos \theta \right)}^2} \\ \Rightarrow - {\sin }^2\theta - 2 \cos \theta - {\cos }^2\theta = 0 \Rightarrow 2 \cos \theta = -1 \\ \Rightarrow \theta =\frac{2\pi }{3} \end{gathered}$$

Therefore, for the above value of θ, the angle between the resultant and vector u is maximum. Therefore, the maximum value of the angle α is,

$$\begin{gathered} \tan \alpha =\frac{{\displaystyle sin \theta }}{{\displaystyle \left(2+ cos \theta \right)}} \Rightarrow tan \alpha =\frac{{\displaystyle sin \left(\frac{2\pi }{3}\right)}}{{\displaystyle \left(2+ cos \left(\frac{2\pi }{3}\right) \right)}} \\ \Rightarrow tan \alpha =\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \Rightarrow tan \alpha =\frac{1}{\sqrt{3}} \Rightarrow \alpha =ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \Rightarrow \alpha =\frac{\pi }{6} \end{gathered}$$
 

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