Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $60^{\circ }$. Then the value of $|\vec{a}-\vec{b}|$ is

Solution

1. Given that $\vec{a}$ and $\vec{b}$ are unit vectors, we have $|\vec{a}|=1$, $|\vec{b}|=1$, and the angle $\theta =60^{\circ }$.
2. The squared magnitude is given by $|\vec{a}-\vec{b}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2-2|\vec{a}||\vec{b}|\cos (60^{\circ })$.
3. Substituting the known values, we get $$\begin{array}{rl}|\vec{a}-\vec{b}{|}^2& =1^2+1^2-2(1)(1)(0.5)\\ & =2-1\\ & =1\end{array}$$.
4. Taking the square root of both sides, we find $|\vec{a}-\vec{b}|=1$.
5. Hence the answer is (A).