If vec{a} and vec{b} are perpendicular unit v | vedclass

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(C) Given that $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors,we have $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $\vec{a} \cdot \vec{b} = 0$.Given $\

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$-1$

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(C) Given that $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors,we have $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $\vec{a} \cdot \vec{b} = 0$.Given $\vec{c} = \vec{a} + \vec{b}$.We know the scalar triple product identity $(\vec{u} 	imes \vec{v}) \cdot (\vec{w} 	imes \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$.Applying this to each term:$1$. $(\vec{a} 	imes \vec{b}) \cdot (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) = 0 \cdot (\vec{b} \cdot \vec{c}) - (\vec{a} \cdot (\vec{a} + \vec{b}))(1) = -(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b}) = -(1 + 0) = -1$.$2$. $(\vec{b} 	imes \vec{c}) \cdot (\vec{c} 	imes \vec{a}) = (\vec{b} \cdot \vec{c})(\vec{c} \cdot \vec{a}) - (\vec{b} \cdot \vec{a})(\vec{c} \cdot \vec{c}) = (\vec{b} \cdot (\vec{a} + \vec{b}))(\vec{c} \cdot \vec{a}) - 0 = (\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b})(\vec{c} \cdot \vec{a}) = (0 + 1)(\vec{c} \cdot \vec{a}) = \vec{c} \cdot \vec{a} = (\vec{a} + \vec{b}) \cdot \vec{a} = 1 + 0 = 1$.$3$. $(\vec{c} 	imes \vec{a}) \cdot (\vec{a} 	imes \vec{b}) = (\vec{c} \cdot \vec{a})(\vec{a} \cdot \vec{b}) - (\vec{c} \cdot \vec{b})(\vec{a} \cdot \vec{a}) = (\vec{c} \cdot \vec{a})(0) - ((\vec{a} + \vec{b}) \cdot \vec{b})(1) = -(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}) = -(0 + 1) = -1$.Summing these values: $-1 + 1 - 1 = -1$.

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