If a and b are unit vectors and a - b is also | vedclass

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Also,$|a - b| = 1$. Squaring both sides of $|a - b| = 1$,we get $|a - b|^2 = 1

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$\frac{\pi}{3}$

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(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Also,$|a - b| = 1$. Squaring both sides of $|a - b| = 1$,we get $|a - b|^2 = 1^2$. Using the property $|x|^2 = x \cdot x$,we have $(a - b) \cdot (a - b) = 1$. Expanding the dot product,we get $a \cdot a - a \cdot b - b \cdot a + b \cdot b = 1$. Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$,the equation becomes $1 - 2(a \cdot b) + 1 = 1$. $2 - 2(a \cdot b) = 1$. $2(a \cdot b) = 1$,which implies $a \cdot b = \frac{1}{2}$. We know that $a \cdot b = |a||b| \cos 	heta$,where $	heta$ is the angle between $a$ and $b$. Substituting the values,$1 	imes 1 	imes \cos 	heta = \frac{1}{2}$. $\cos 	heta = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$ or $60^\circ$.

If $a$ and $b$ are unit vectors and $a - b$ is also a unit vector,then the angle between $a$ and $b$ is

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