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Question Number 187496 by mathlove last updated on 18/Feb/23

Commented by mr W last updated on 18/Feb/23

$$thingslike\mid \frac{\overrightarrow{A}}{\overrightarrow{B}}\mid {don}^{\prime }texist!$$$$ifyoumean\frac{\mid \overrightarrow{A}\mid }{\mid \overrightarrow{B}\mid },thenyoushould$$$$alsowritethatwhatyoumean.$$

Answered by manxsol last updated on 18/Feb/23

$$\frac{\mid p-q{\mid }^2}{\mid p+q{\mid }^2}=\frac{\mid p{\mid }^2-2p\bullet q+\mid q{\mid }^2}{\mid p{\mid }^2+2p\bullet q+\mid q{\mid }^2}$$$$=\frac{2-2cos\theta }{2+2cos\theta }p\bullet q=\left(1\right)\left(1\right)cos\theta$$$$\mid \frac{p-q}{p+q}{\mid }^2=\frac{1-cos\theta }{1+cos\theta }=\frac{1-cos\frac{\theta }{2}}{1+cos\frac{\theta }{2}}$$$$=\frac{1-(1-2{sin}^2\frac{\theta }{2})}{1+(2{cos}^2\frac{\theta }{2}-1)}=\frac{2{sin}^2\frac{\theta }{2}}{2{cos}^2\frac{\theta }{2}}$$$$\mid \frac{p-q}{p+q}\mid =tan\frac{\theta }{2}$$$$$$$$$$

Commented by manxsol last updated on 18/Feb/23

$$true,noexistedivision$$$$devectores$$

Commented by mr W last updated on 18/Feb/23

$$¡Denada!$$

Commented by manxsol last updated on 18/Feb/23

$$exercisedebeserasi$$$$toprove\frac{\mid \overrightarrow{p}-\overrightarrow{q}{\mid }^2}{\mid \overrightarrow{p}+\overrightarrow{q}{\mid }^2}=tg\frac{\theta }{2}$$$$pandqvectorsunitary$$$$$$$$graciasporlaatencion$$$$SirW$$

Commented by mr W last updated on 18/Feb/23

$$whatis\mid \frac{p-q}{p+q}\mid ?$$

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