Let $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ be three unit vectors such that $$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c})$$. If $$\vec{b}$$ is not parallel to $$\vec{c}$$, then the angle between $$\vec{a}$$ and $$\vec{b}$$ is

We are told that $$\vec a,\;\vec b,\;\vec c$$ are unit vectors (that is, each has magnitude 1) and that they satisfy the vector equation

$$\vec a \times (\vec b \times \vec c)=\frac{\sqrt3}{2}\,(\vec b+\vec c).$$

Because $$\vec b$$ is not parallel to $$\vec c$$, we may treat $$\vec b$$ and $$\vec c$$ as independent directions. The key tool here is the standard vector triple-product formula:

$$\boxed{\;\vec p \times (\vec q \times \vec r)=\vec q(\vec p\cdot\vec r)\;-\;\vec r(\vec p\cdot\vec q)\;}.$$

We apply this identity with $$\vec p=\vec a,\;\vec q=\vec b,\;\vec r=\vec c$$. Hence

$$\vec a \times (\vec b \times \vec c) =\vec b\,(\vec a\cdot\vec c)\;-\;\vec c\,(\vec a\cdot\vec b).$$

To keep the algebra transparent, we introduce the scalar symbols

$$x=\vec a\cdot\vec c,\qquad y=\vec a\cdot\vec b.$$

Both $$x$$ and $$y$$ are simply cosines of the respective angles, because each pair of vectors involved has unit length. Substituting these into the expression for the triple product gives

$$\vec a \times (\vec b \times \vec c)=x\,\vec b\;-\;y\,\vec c.$$

But the problem statement tells us that the very same vector equals $$\tfrac{\sqrt3}{2}\,(\vec b+\vec c)$$. Therefore we equate the two expressions:

$$x\,\vec b\;-\;y\,\vec c=\frac{\sqrt3}{2}\,\vec b+\frac{\sqrt3}{2}\,\vec c.$$

Since $$\vec b$$ and $$\vec c$$ are not parallel, the only way two linear combinations of them can be equal is for the coefficients of $$\vec b$$ to match and, independently, the coefficients of $$\vec c$$ to match. Thus we obtain the pair of equations

$$x=\frac{\sqrt3}{2},\qquad -y=\frac{\sqrt3}{2}.$$

From the second equation we get

$$y=-\,\frac{\sqrt3}{2}.$$

Recall now that $$y=\vec a\cdot\vec b$$, and for unit vectors the dot product equals the cosine of the angle $$\theta$$ between them: $$\vec a\cdot\vec b=\cos\theta$$. Hence

$$\cos\theta=-\,\frac{\sqrt3}{2}.$$

The cosine value $$-\frac{\sqrt3}{2}$$ corresponds to an angle of $$150^\circ$$ (or $$5\pi/6$$ radians) in the principal range $$0\le\theta\le\pi$$.

Therefore, the angle between $$\vec a$$ and $$\vec b$$ is

$$\theta=\frac{5\pi}{6}.$$

Hence, the correct answer is Option B.