JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 7)
If two vectors $$\vec{A}$$ and $$\vec{B}$$ having equal magnitude $$R$$ are inclined at angle $$\theta$$, then
$$|\vec{A}+\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)$$
$$|\vec{A}-\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)$$
$$|\vec{A}-\vec{B}|=\sqrt{2} R \sin \left(\frac{\theta}{2}\right)$$
$$|\vec{A}+\vec{B}|=2 R \sin \left(\frac{\theta}{2}\right)$$
Explanation
The magnitude of resultant vector
$$R^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}$$
Here $$a=b=R$$
Then $$R^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}$$
$$\begin{aligned} & =R \sqrt{2} \sqrt{1+\cos \theta} \\ & =\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}} \\ & =2 R \cos \frac{\theta}{2} \end{aligned}$$
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