If hat{a}, hat{b} and hat{c} are unit vectors | vedclass

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(B) Given that $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,so $|\hat{a}| = |\hat{b}| = |\hat{c}| = 1$. Given $\hat{a}+\hat{b}+\hat{c}=\vec{0}$. Takin

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$-\frac{3}{2}$

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(B) Given that $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,so $|\hat{a}| = |\hat{b}| = |\hat{c}| = 1$. Given $\hat{a}+\hat{b}+\hat{c}=\vec{0}$. Taking the dot product of the sum with itself: $(\hat{a}+\hat{b}+\hat{c}) \cdot (\hat{a}+\hat{b}+\hat{c}) = \vec{0} \cdot \vec{0} = 0$. Expanding the dot product: $\hat{a} \cdot \hat{a} + \hat{b} \cdot \hat{b} + \hat{c} \cdot \hat{c} + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 0$. Since $|\hat{a}|^2 = \hat{a} \cdot \hat{a} = 1$,we have: $1 + 1 + 1 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 0$. $3 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 0$. $2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = -3$. Therefore,$\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a} = -\frac{3}{2}$.

If $\hat{a}, \hat{b}$ and $\hat{c}$ are unit vectors such that $\hat{a}+\hat{b}+\hat{c}=\vec{0}$,then the value of $\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}$ is equal to

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