If |a| = |b|,then (a + b) cdot (a - b) is: | vedclass

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(C) Given that $|a| = |b|$.We need to evaluate the dot product $(a + b) \cdot (a - b)$.Using the distributive property of the dot product:$(a + b) \cd

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(C) Given that $|a| = |b|$.We need to evaluate the dot product $(a + b) \cdot (a - b)$.Using the distributive property of the dot product:$(a + b) \cdot (a - b) = a \cdot a - a \cdot b + b \cdot a - b \cdot b$Since the dot product is commutative $(a \cdot b = b \cdot a)$,the terms $-a \cdot b$ and $b \cdot a$ cancel each other out.Thus,$(a + b) \cdot (a - b) = a \cdot a - b \cdot b$.We know that $a \cdot a = |a|^2$ and $b \cdot b = |b|^2$.So,$(a + b) \cdot (a - b) = |a|^2 - |b|^2$.Since $|a| = |b|$,it follows that $|a|^2 = |b|^2$.Therefore,$|a|^2 - |b|^2 = 0$.

If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

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