If $\vec a$ and $\vec b$ are two unit vectors and $\vec a+\vec b$ is also unit vector, the magnitude of $\vec a-\vec b$ is

$\sqrt{3}$

The correct answer is Option (4) → $\sqrt{3}$

Given: $\vec{a}$ and $\vec{b}$ are unit vectors, and $\vec{a} + \vec{b}$ is also a unit vector.

So, $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $|\vec{a} + \vec{b}| = 1$.

Use identity: $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$

$1^2 = 1^2 + 1^2 + 2\vec{a} \cdot \vec{b}$

$1 = 1 + 1 + 2\vec{a} \cdot \vec{b} \Rightarrow 2\vec{a} \cdot \vec{b} = -1$

$\vec{a} \cdot \vec{b} = -\frac{1}{2}$

Now compute $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$

$= 1 + 1 - 2(-\frac{1}{2}) = 2 + 1 = 3$

So, $|\vec{a} - \vec{b}| = \sqrt{3}$