Find |vec{a}-vec{b}|,if two vectors vec{a} an | vedclass

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(A) We know that $|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.Expanding the dot product,we get:$|\vec{a} - \vec{b}|^2 = \ve

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$\sqrt{5}$

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(A) We know that $|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.Expanding the dot product,we get:$|\vec{a} - \vec{b}|^2 = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$.Using the properties $\vec{a} \cdot \vec{a} = |\vec{a}|^2$,$\vec{b} \cdot \vec{b} = |\vec{b}|^2$,and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,we have:$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$.Substituting the given values $|\vec{a}| = 2$,$|\vec{b}| = 3$,and $\vec{a} \cdot \vec{b} = 4$:$|\vec{a} - \vec{b}|^2 = (2)^2 - 2(4) + (3)^2$.$|\vec{a} - \vec{b}|^2 = 4 - 8 + 9$.$|\vec{a} - \vec{b}|^2 = 5$.Therefore,$|\vec{a} - \vec{b}| = \sqrt{5}$.

Find $|\vec{a}-\vec{b}|$,if two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2, |\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$.

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