For any two non-zero vectors vec{a} and vec{b | vedclass

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(B) We are given the expression $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.Using the distributive property of the dot product,we expand t

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(B) We are given the expression $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.Using the distributive property of the dot product,we expand the expression:$= (a \vec{b}) \cdot (a \vec{b}) - (a \vec{b}) \cdot (b \vec{a}) + (b \vec{a}) \cdot (a \vec{b}) - (b \vec{a}) \cdot (b \vec{a})$$= a^2 (\vec{b} \cdot \vec{b}) - ab (\vec{b} \cdot \vec{a}) + ab (\vec{a} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$Since the dot product is commutative,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,so the middle terms cancel out:$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$Wait,let us re-evaluate the expression given in the prompt: $(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$.$= a^2 (\vec{b} \cdot \vec{b}) - b^2 (\vec{a} \cdot \vec{a})$$= a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2$If $a$ and $b$ are scalars representing the magnitudes $|\vec{a}|$ and $|\vec{b}|$,then $a^2 |\vec{b}|^2 - b^2 |\vec{a}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2 = 0$.

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

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