If $|\vec a-\vec r||\vec a|=|\vec r|= 1$, then angle between $\vec a$ and $\vec r$ is

$\frac{\pi}{3}$

The correct answer is Option (1) → $\frac{\pi}{3}$

Given:

$|\vec{a} - \vec{r}| \cdot |\vec{a}| = |\vec{r}| = 1$

Since $|\vec{a}| = 1$ and $|\vec{r}| = 1$, the condition becomes:

$|\vec{a} - \vec{r}| = 1$

Now expand using vector identity:

$|\vec{a} - \vec{r}|^2 = (\vec{a} - \vec{r}) \cdot (\vec{a} - \vec{r}) = |\vec{a}|^2 + |\vec{r}|^2 - 2\vec{a} \cdot \vec{r}$

Since $|\vec{a}| = |\vec{r}| = 1$, this becomes:

$|\vec{a} - \vec{r}|^2 = 1^2 + 1^2 - 2\cos\theta = 2 - 2\cos\theta$

Also, from given condition: $|\vec{a} - \vec{r}| = 1 \Rightarrow |\vec{a} - \vec{r}|^2 = 1$

So equating:

$2 - 2\cos\theta = 1$

$\Rightarrow \cos\theta = \frac{1}{2}$

$\Rightarrow \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$