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} = \overrighta

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 $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$ has a magnitude $R = \sqrt{a^2 + a^2 + 2a^2 \cos 	heta} = \sqrt{2a^2(1 + \cos 	heta)} = \sqrt{2a^2(2 \cos^2 \frac{	heta}{2})} = 2a \cos \frac{	heta}{2}$. The angle $\alpha$ that the resultant $\overrightarrow{R}$ makes with vector $\overrightarrow{A}$ is given by $	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta}$. Substituting $A = B = a$,we get $	an \alpha = \frac{a \sin 	heta}{a + a \cos 	heta} = \frac{\sin 	heta}{1 + \cos 	heta} = \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}$. Since the magnitudes are equal,the resultant bisects the angle between the two vectors.

$\overrightarrow{A}$ 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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