Let $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ be three unit vectors, out of which vectors $$\vec{b}$$ and $$\vec{c}$$ are non-parallel. If $$\alpha$$ and $$\beta$$ are the angles which vector $$\vec{a}$$ makes with vectors $$\vec{b}$$ and $$\vec{c}$$ respectively and $$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}\vec{b}$$, then $$|\alpha - \beta|$$ is equal to:

We are given three unit vectors $$\vec a,\;\vec b,\;\vec c$$ with the information that $$\vec b$$ and $$\vec c$$ are not parallel. The angles that $$\vec a$$ makes with $$\vec b$$ and $$\vec c$$ are denoted by $$\alpha$$ and $$\beta$$ respectively, so by definition of the dot product for unit vectors we have

$$\vec a\cdot\vec b=\cos\alpha\qquad\text{and}\qquad\vec a\cdot\vec c=\cos\beta.$$

We are also told that

$$\vec a\times(\vec b\times\vec c)=\dfrac12\,\vec b.$$

To proceed, we recall the vector triple-product identity. For any vectors $$\vec p,\vec q,\vec r$$ the identity states

$$\vec p\times(\vec q\times\vec r)=\vec q\,(\vec p\cdot\vec r)-\vec r\,(\vec p\cdot\vec q).$$

Applying this identity with $$\vec p=\vec a,\;\vec q=\vec b,\;\vec r=\vec c$$, we obtain

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

Using the shorthand $$\vec a\cdot\vec b=\cos\alpha$$ and $$\vec a\cdot\vec c=\cos\beta$$, the expression becomes

$$\vec a\times(\vec b\times\vec c)=\vec b\,\cos\beta-\vec c\,\cos\alpha.$$

But the problem statement gives the same vector as $$\dfrac12\,\vec b$$. Hence we equate the two results:

$$\vec b\,\cos\beta-\vec c\,\cos\alpha=\dfrac12\,\vec b.$$

Because $$\vec b$$ and $$\vec c$$ are non-parallel, they are linearly independent directions. Therefore the coefficients of $$\vec b$$ and $$\vec c$$ on both sides must match **individually**.

Comparing the $$\vec b$$ components:

$$\cos\beta=\dfrac12.$$

Comparing the $$\vec c$$ components (there is no $$\vec c$$ term on the right-hand side, so its coefficient is zero):

$$-\cos\alpha=0\;\;\Longrightarrow\;\;\cos\alpha=0.$$

Since all three vectors are unit vectors, the cosine values directly yield the angles:

• From $$\cos\alpha=0$$ we find $$\alpha=90^\circ.$$ • From $$\cos\beta=\dfrac12$$ we find $$\beta=60^\circ.$$

Hence the absolute difference between the two angles is

$$|\alpha-\beta|=\bigl|90^\circ-60^\circ\bigr|=30^\circ.$$

Hence, the correct answer is Option 4.