overrightarrow{A} and overrightarrow{B} are t | vedclass

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(B) Let the magnitudes of the two vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = a$.The resultant vector $\overrightarrow{R} = \overrightar

Vedclass · Accepted Answer

$\frac{	heta}{2}$

Vedclass · Answer

(B) Let the magnitudes of the two vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = a$.The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$ makes an angle $\alpha$ with vector $\overrightarrow{A}$, given by the formula:$	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta}$Since $A = B = a$, we have:$	an \alpha = \frac{a \sin 	heta}{a + a \cos 	heta} = \frac{\sin 	heta}{1 + \cos 	heta}$Using trigonometric identities $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$ and $1 + \cos 	heta = 2 \cos^2(\frac{	heta}{2})$:$	an \alpha = \frac{2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})}{2 \cos^2(\frac{	heta}{2})} = 	an(\frac{	heta}{2})$Therefore, $\alpha = \frac{	heta}{2}$.This shows that the resultant vector bisects the angle between the two vectors of equal magnitude.

$\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors of equal magnitudes and $\theta$ is the angle between them. The angle between $\overrightarrow{A}$ or $\overrightarrow{B}$ with their resultant is

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