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(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Since $a+b$ is a unit vector,$|a+b| = 1$. Squaring both sides,we get $|a+b|^2

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$3$

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(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Since $a+b$ is a unit vector,$|a+b| = 1$. Squaring both sides,we get $|a+b|^2 = 1^2 = 1$. Using the identity $|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have $1 + 1 + 2(a \cdot b) = 1$,which implies $2(a \cdot b) = -1$,so $a \cdot b = -1/2$. Now,we need to find $|a-b|^2$. Using the identity $|a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$,we substitute the known values: $|a-b|^2 = 1^2 + 1^2 - 2(-1/2) = 1 + 1 + 1 = 3$.

If $a$ and $b$ are two unit vectors such that $a+b$ is also a unit vector,then $|a-b|^2=$

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