If $$\vec{a}$$ and $$\vec{b}$$ are unit vectors, then the greatest value of $$\sqrt{3}|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$$ is___.

Let $$\vec a$$ and $$\vec b$$ be unit vectors. Hence we have $$|\vec a| = 1$$ and $$|\vec b| = 1$$. Let the angle between the two vectors be $$\theta$$, so that $$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = \cos\theta$$.

First we write the standard formula for the magnitude of the sum of two vectors:

$$|\vec a+\vec b|^2 = |\vec a|^2 + |\vec b|^2 + 2\vec a\cdot\vec b.$$

Substituting $$|\vec a|=1,\;|\vec b|=1,\;\vec a\cdot\vec b=\cos\theta$$ we get

$$|\vec a+\vec b|^2 = 1 + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta).$$

Taking the square root,

$$|\vec a+\vec b| = \sqrt{2(1+\cos\theta)}.$$

Exactly in the same way, for the difference we use

$$|\vec a-\vec b|^2 = |\vec a|^2 + |\vec b|^2 - 2\vec a\cdot\vec b.$$

Again substituting the same values,

$$|\vec a-\vec b|^2 = 1 + 1 - 2\cos\theta = 2 - 2\cos\theta = 2(1-\cos\theta).$$

Therefore

$$|\vec a-\vec b| = \sqrt{2(1-\cos\theta)}.$$

Now we simplify each square root by expressing $$1\pm\cos\theta$$ in half-angle form. We recall the trigonometric identities

$$1+\cos\theta = 2\cos^2\frac\theta2,\qquad 1-\cos\theta = 2\sin^2\frac\theta2.$$

Substituting these into the magnitudes we get

$$|\vec a+\vec b| = \sqrt{2\cdot 2\cos^2\frac\theta2} = \sqrt{4\cos^2\frac\theta2} = 2\left|\cos\frac\theta2\right|,$$

$$|\vec a-\vec b| = \sqrt{2\cdot 2\sin^2\frac\theta2} = \sqrt{4\sin^2\frac\theta2} = 2\left|\sin\frac\theta2\right|.$$

Because the absolute value bars already appear, we may let $$\phi=\dfrac{\theta}{2}$$ and consider $$\phi\in[0,\pi]$$. In the interval $$[0,\tfrac\pi2]$$ both sine and cosine are non-negative, and by symmetry the same maximum is obtained there, so we may drop the absolute values for maximisation. Setting

$$f(\phi)=\sqrt3\,|\vec a+\vec b|+|\vec a-\vec b| = \sqrt3\,(2\cos\phi)+2\sin\phi = 2\bigl(\sqrt3\cos\phi+\sin\phi\bigr),$$

we need to maximise

$$g(\phi)=\sqrt3\cos\phi+\sin\phi\quad\text{for }0\le\phi\le\dfrac\pi2.$$

We differentiate $$g(\phi)$$:

$$\dfrac{dg}{d\phi} = -\sqrt3\sin\phi + \cos\phi.$$

The critical point is obtained by setting the derivative equal to zero:

$$-\sqrt3\sin\phi + \cos\phi = 0 \;\Longrightarrow\; \cos\phi = \sqrt3\sin\phi \;\Longrightarrow\; \tan\phi = \dfrac1{\sqrt3}.$$

Thus

$$\phi = \tan^{-1}\!\left(\dfrac1{\sqrt3}\right) = \dfrac\pi6.$$

To confirm that this point is a maximum, we compute the second derivative:

$$\dfrac{d^2g}{d\phi^2} = -\sqrt3\cos\phi - \sin\phi,$$

and substituting $$\phi=\dfrac\pi6$$ gives

$$\dfrac{d^2g}{d\phi^2}\Big|_{\phi=\pi/6} = -\sqrt3\left(\dfrac{\sqrt3}{2}\right) - \dfrac12 = -\dfrac32 - \dfrac12 = -2 < 0,$$

so the function indeed attains a maximum there.

Evaluating $$g(\phi)$$ at $$\phi=\dfrac\pi6$$:

$$g\!\left(\dfrac\pi6\right) = \sqrt3\cos\!\left(\dfrac\pi6\right) + \sin\!\left(\dfrac\pi6\right) = \sqrt3\left(\dfrac{\sqrt3}{2}\right) + \dfrac12 = \dfrac32 + \dfrac12 = 2.$$

Therefore the maximum value of the original expression is

$$f_{\max} = 2 \times g_{\max} = 2 \times 2 = 4.$$

Hence, the correct answer is Option 4.