The vectors $\vec{a}$ and $\vec{b}$ each of magnitude $x$ are inclined to each other such that their resultant is equal to $\sqrt{3}x$. Then magnitude of the resultant of $\vec{a}$ and $-\vec{b}$ is

Solution:

Let the magnitude of vectors $\vec{a}$ and $\vec{b}$ be $|\vec{a}| = x$ and $|\vec{b}| = x$.

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.

The magnitude of the resultant of $\vec{a}$ and $\vec{b}$ is given by the formula:

$|\vec{R}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta}$

We are given that $|\vec{R}| = \sqrt{3}x$.

Substituting the given values:

$(\sqrt{3}x)^2 = x^2 + x^2 + 2(x)(x)\cos\theta$

$3x^2 = 2x^2 + 2x^2\cos\theta$

Subtract $2x^2$ from both sides:

$x^2 = 2x^2\cos\theta$

Divide by $2x^2$ (assuming $x \neq 0$):

$\cos\theta = \frac{x^2}{2x^2} = \frac{1}{2}$

This implies that $\theta = 60^\circ$.

Now, we need to find the magnitude of the resultant of $\vec{a}$ and $-\vec{b}$.

The angle between $\vec{a}$ and $-\vec{b}$ will be $180^\circ - \theta = 180^\circ - 60^\circ = 120^\circ$.

Let the new resultant be $\vec{R}' = \vec{a} + (-\vec{b})$.

The magnitude is:

$|\vec{R}'| = \sqrt{|\vec{a}|^2 + |-\vec{b}|^2 + 2|\vec{a}||-\vec{b}|\cos(180^\circ - \theta)}$

Since $|-\vec{b}| = |\vec{b}| = x$, we have:

$|\vec{R}'| = \sqrt{x^2 + x^2 + 2(x)(x)\cos(120^\circ)}$

We know that $\cos(120^\circ) = -\frac{1}{2}$.

$|\vec{R}'| = \sqrt{2x^2 + 2x^2(-\frac{1}{2})}$

$|\vec{R}'| = \sqrt{2x^2 - x^2}$

$|\vec{R}'| = \sqrt{x^2}$

$|\vec{R}'| = x$

Thus, the magnitude of the resultant of $\vec{a}$ and $-\vec{b}$ is $x$. This corresponds to option (3).