If the sum of two unit vectors is a unit vect | vedclass

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(B) Let the two unit vectors be $\vec{A}$ and $\vec{B}$,where $|\vec{A}| = 1$ and $|\vec{B}| = 1$.Given that their sum is also a unit vector: $|\vec{A

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$\sqrt{3}$

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(B) Let the two unit vectors be $\vec{A}$ and $\vec{B}$,where $|\vec{A}| = 1$ and $|\vec{B}| = 1$.Given that their sum is also a unit vector: $|\vec{A} + \vec{B}| = 1$.Squaring both sides: $|\vec{A} + \vec{B}|^2 = 1^2$.$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta = 1$,where $	heta$ is the angle between them.$1 + 1 + 2(1)(1) \cos 	heta = 1 \implies 2 + 2 \cos 	heta = 1 \implies 2 \cos 	heta = -1 \implies \cos 	heta = -1/2$.Now,we need to find the magnitude of their difference: $|\vec{A} - \vec{B}|$.$|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos 	heta$.Substituting the values: $|\vec{A} - \vec{B}|^2 = 1 + 1 - 2(1)(1)(-1/2)$.$|\vec{A} - \vec{B}|^2 = 2 + 1 = 3$.Therefore,$|\vec{A} - \vec{B}| = \sqrt{3}$.

If the sum of two unit vectors is a unit vector,find the magnitude of their difference.

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