If the sum of two unit vectors is also a unit | 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 $\vec{R} = \vec{A} + \vec{B}$ i

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$\sqrt{3}, 120^{\circ}$

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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 $\vec{R} = \vec{A} + \vec{B}$ is also a unit vector,so $|\vec{R}| = 1$.The magnitude of the resultant vector is given by $|\vec{R}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta} = 1$.Squaring both sides: $1^2 + 1^2 + 2(1)(1) \cos 	heta = 1^2$.$2 + 2 \cos 	heta = 1 \Rightarrow 2 \cos 	heta = -1 \Rightarrow \cos 	heta = -1/2$.Thus,the angle $	heta = 120^{\circ}$.Now,the magnitude of their difference is $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 	heta}$.Substituting the values: $|\vec{A} - \vec{B}| = \sqrt{1^2 + 1^2 - 2(1)(1) \cos 120^{\circ}}$.Since $\cos 120^{\circ} = -1/2$,we get $|\vec{A} - \vec{B}| = \sqrt{1 + 1 - 2(-1/2)} = \sqrt{1 + 1 + 1} = \sqrt{3}$.Therefore,the magnitude of the difference is $\sqrt{3}$ and the angle is $120^{\circ}$.

If the sum of two unit vectors is also a unit vector,then the magnitude of their difference and the angle between the two given unit vectors is ..............

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