If |vec a| = 1,|vec b|= 2, |2vec a +vec b| = 2√3 then |vec a-vec b| i…

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

Given:

$|\vec{a}| = 1$,

$|\vec{b}| = 2$,

$|2\vec{a} + \vec{b}| = 2\sqrt{3}$

Find $|\vec{a} - \vec{b}|$.

Use the formula for magnitude squared:

$|2\vec{a} + \vec{b}|^2 = (2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b}) = 4|\vec{a}|^2 + 4 \vec{a} \cdot \vec{b} + |\vec{b}|^2$

Given $|2\vec{a} + \vec{b}| = 2\sqrt{3}$, so:

$(2\sqrt{3})^2 = 4 \times 1^2 + 4 \vec{a} \cdot \vec{b} + 2^2$

$4 \times 3 = 4 + 4 \vec{a} \cdot \vec{b} + 4$

$12 = 8 + 4 \vec{a} \cdot \vec{b}$

$4 \vec{a} \cdot \vec{b} = 12 - 8 = 4$

$\vec{a} \cdot \vec{b} = 1$

Now find $|\vec{a} - \vec{b}|^2$:

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

$= 1 - 2 \times 1 + 4 = 1 - 2 + 4 = 3$

Therefore:

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