If |a| = 1, |b| = 2, |2a + b| = 2√3 then |a - b| is:

The correct answer is Option (1) → \(\sqrt{3}\)

Since \(|\vec{p}|^2 = \vec{p}\cdot\vec{p}\), the given modulus fixes the dot product \(\vec{a}\cdot\vec{b}\):

    \(|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + 4\,\vec{a}\cdot\vec{b} + |\vec{b}|^2\)
    \(12 = 4 + 4\,\vec{a}\cdot\vec{b} + 4\)
    \(\vec{a}\cdot\vec{b} = 1\)

Then

    \(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2\)
    \(= 1 - 2 + 4 = 3\)

So \(|\vec{a} - \vec{b}| = \sqrt{3}\).