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(A) Given that $|a|=1$ and $|b|=1$, and $\alpha$ is the angle between $a$ and $b$. We know that $a \cdot b = |a||b| \cos \alpha = (1)(1) \cos \alpha =

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$-\frac{1}{2}$

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(A) Given that $|a|=1$ and $|b|=1$, and $\alpha$ is the angle between $a$ and $b$. We know that $a \cdot b = |a||b| \cos \alpha = (1)(1) \cos \alpha = \cos \alpha$. Since $a+b$ is a unit vector, we have $|a+b|=1$. Squaring both sides, we get $|a+b|^2 = 1^2 = 1$. Expanding the dot product, $(a+b) \cdot (a+b) = 1$. $a \cdot a + a \cdot b + b \cdot a + b \cdot b = 1$. Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$, and $a \cdot b = b \cdot a = \cos \alpha$, we substitute these values: $1 + \cos \alpha + \cos \alpha + 1 = 1$. $2 + 2 \cos \alpha = 1$. $2 \cos \alpha = 1 - 2$. $2 \cos \alpha = -1$. $\cos \alpha = -\frac{1}{2}$.

If $a$ and $b$ are unit vectors and $\alpha$ is the angle between them, then $a+b$ is a unit vector when $\cos \alpha=$

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