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(A) Let the two vectors be $\vec{P}$ and $\vec{Q}$ such that $|\vec{P}| = |\vec{Q}| = A$.The magnitude of the resultant vector $\vec{R} = \vec{P} + \v

Vedclass · Accepted Answer

$2\,A \cos \frac{	heta}{2}$,along bisector

Vedclass · Answer

(A) Let the two vectors be $\vec{P}$ and $\vec{Q}$ such that $|\vec{P}| = |\vec{Q}| = A$.The magnitude of the resultant vector $\vec{R} = \vec{P} + \vec{Q}$ is given by $R = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$.Substituting the values,we get $R = \sqrt{A^2 + A^2 + 2A^2 \cos 	heta} = \sqrt{2A^2(1 + \cos 	heta)}$.Using the trigonometric identity $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we get $R = \sqrt{2A^2 \cdot 2 \cos^2 \frac{	heta}{2}} = \sqrt{4A^2 \cos^2 \frac{	heta}{2}} = 2A \cos \frac{	heta}{2}$.Since the magnitudes of the two vectors are equal,the resultant vector bisects the angle between them. Thus,the direction is along the angle bisector.

Two vectors having equal magnitudes $A$ make an angle $\theta$ with each other. The magnitude and direction of the resultant are respectively

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