The magnitude of the sum of the two vectors v | vedclass

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(C) Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively.The magnitude of their sum is given by:$|\vec{A}+\vec{B}|

Vedclass · Accepted Answer

$90$

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(C) Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively.The magnitude of their sum is given by:$|\vec{A}+\vec{B}| = \sqrt{A^{2}+B^{2}+2AB \cos 	heta}$,where $	heta$ is the angle between the vectors.The magnitude of their difference is given by:$|\vec{A}-\vec{B}| = \sqrt{A^{2}+B^{2}-2AB \cos 	heta}$.Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides:$A^{2}+B^{2}+2AB \cos 	heta = A^{2}+B^{2}-2AB \cos 	heta$.Subtracting $A^{2}+B^{2}$ from both sides,we get:$2AB \cos 	heta = -2AB \cos 	heta$.$4AB \cos 	heta = 0$.Since $A$ and $B$ are magnitudes of vectors,$A 
eq 0$ and $B 
eq 0$,therefore $\cos 	heta = 0$.This implies $	heta = 90^{\circ}$.

The magnitude of the sum of the two vectors $\vec{A}$ and $\vec{B}$ is equal to the magnitude of the difference of the two vectors $\vec{A}$ and $\vec{B}$. The angle between $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

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