Let \(\vec{a},\vec{b} ,(\vec{a}\times\vec{b})\) be unit vectors. What is \(\vec{a}.\vec{b}\)$\stackrel{equal\; to?}{}$

Calculation:

Given,

The vectors \(\vec{a},\vec{b} ,(\vec{a}\times\vec{b})\) are unit vectors.

Since \(\vec a\) and  \(\vec{b}\) are unit vectors, we know:

\( |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \).

The magnitude of the cross product \(( \vec{a} \times \vec{b} )\) is given by:

\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \times 1 \times \sin \theta = \sin \theta \).

Since \( ( |\vec{a} \times \vec{b}| = 1 \), we have:

\( \sin \theta = 1 \), so \(\theta = 90^\circ \), meaning \(\vec a \) and \( \vec{b} \) are perpendicular.

The dot product \( ( \vec{a} \cdot \vec{b} )\) is:

\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \times 1 \times \cos 90^\circ = 0 \).

∴ The value of\(( \vec{a} \cdot \vec{b} ) \) is 0.

Hence, the correct answer is Option 1.