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સમાન મૂલ્ય $R$ ધરાવતા બે સદીશો $\vec{A}$ અને $\vec{B}$ વચ્ચેનો ખૂણો $\theta$ છે તો
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{2} \mathrm{R} \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=2 \mathrm{R} \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=2 \mathrm{R} \cos \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=2 R \cos \left(\frac{\theta}{2}\right)$
Solution
The magnitude of resultant vector
$R^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}$
Here $a=b=R$
Then $R^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}$
$=R \sqrt{2} \sqrt{1+\cos \theta}$
$=\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}}$
$=2 R \cos \frac{\theta}{2}$
