Suppose the angle between two nonzero vectors A and B is 120^ ^ ^ ,...

The magnitude of the resultant vector $\vec{C}$ is given by:

$C = |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$

Given $\theta = 120^{{}^{{}^{\circ}}}$, $\cos 120^{{}^{{}^{\circ}}} = -\dfrac{1}{2}$. Substituting this value:

$C = \sqrt{A^2 + B^2 - AB}$

The expression for $|A - B|$ can be written as:

$|A - B| = \sqrt{(A - B)^2} = \sqrt{A^2 + B^2 - 2AB}$

Since $\vec{A}$ and $\vec{B}$ are nonzero vectors, their magnitudes $A$ and $B$ are strictly positive, which means $AB > 0$.

Comparing the terms inside the square roots:

$A^2 + B^2 - AB > A^2 + B^2 - 2AB$

Taking the square root of both sides yields:

$\sqrt{A^2 + B^2 - AB} > \sqrt{A^2 + B^2 - 2AB}$

$C > |A - B|$

Thus, $C$ is necessarily greater than $|A - B|$.