If |overrightarrow{A}|=4 unit,|overrightarrow | vedclass

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(A) Given: $|\vec{A}|=4$,$|\vec{A}+\vec{B}|=10$,and $\vec{A} \cdot(\vec{A}+\vec{B})=20$.Expanding the dot product: $\vec{A} \cdot \vec{A} + \vec{A} \c

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$\sqrt{76}$ units

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(A) Given: $|\vec{A}|=4$,$|\vec{A}+\vec{B}|=10$,and $\vec{A} \cdot(\vec{A}+\vec{B})=20$.Expanding the dot product: $\vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} = 20$.Since $\vec{A} \cdot \vec{A} = |\vec{A}|^2 = 4^2 = 16$,we have $16 + \vec{A} \cdot \vec{B} = 20$,which implies $\vec{A} \cdot \vec{B} = 4$.Now,consider the magnitude of the sum: $|\vec{A}+\vec{B}|^2 = (\vec{A}+\vec{B}) \cdot (\vec{A}+\vec{B}) = |\vec{A}|^2 + |\vec{B}|^2 + 2(\vec{A} \cdot \vec{B})$.Substituting the known values: $10^2 = 4^2 + |\vec{B}|^2 + 2(4)$.$100 = 16 + |\vec{B}|^2 + 8$.$100 = 24 + |\vec{B}|^2$.$|\vec{B}|^2 = 100 - 24 = 76$.Therefore,$|\vec{B}| = \sqrt{76}$ units.

If $|\overrightarrow{A}|=4$ unit,$|\overrightarrow{A}+\overrightarrow{B}|=10$ unit and $\overrightarrow{A} \cdot(\overrightarrow{A}+\overrightarrow{B})=20$ unit,then $|\overrightarrow{B}|=$ ?

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