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(D) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$.Using the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (a

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$90^\circ, 60^\circ$

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(D) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$.Using the vector triple product formula: $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$.Given $a 	imes (b 	imes c) = \frac{1}{2}b$,we have $(a \cdot c)b - (a \cdot b)c = \frac{1}{2}b$.Comparing the coefficients of $b$ and $c$ (assuming $b$ and $c$ are non-collinear),we get:$a \cdot c = \frac{1}{2}$ and $a \cdot b = 0$.For the angle $	heta_1$ between $a$ and $b$: $\cos 	heta_1 = \frac{a \cdot b}{|a||b|} = 0 \implies 	heta_1 = 90^\circ$.For the angle $	heta_2$ between $a$ and $c$: $\cos 	heta_2 = \frac{a \cdot c}{|a||c|} = \frac{1/2}{1 	imes 1} = \frac{1}{2} \implies 	heta_2 = 60^\circ$.Thus,the angles are $90^\circ$ and $60^\circ$ respectively.

If three unit vectors $a, b, c$ are such that $a \times (b \times c) = \frac{b}{2},$ then the vector $a$ makes with $b$ and $c$ respectively the angles

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