The position vectors of three points A, B and C respectively, where  \(\vec{a} ,\vec{b} \) and \(\vec{c} \)  respectively, where \(\vec{c} = (\cos^2 \theta)\vec{a}+(\sin^2 \theta)\vec{b}\). What is \((\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})\) equal to?

Calculation:

Given,

The position vectors of points A, B, and C are \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) respectively, and \( \vec{c} = \cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b} \).

The expression to evaluate is: \( (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) \).

First, substitute \( \vec{c} \) into the equation:

\( (\vec{a} \times \vec{b}) + (\vec{b} \times (\cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b})) + (\cos^2 \theta \, \vec{a} + \sin^2 \theta \, \vec{b}) \times \vec{a} \).

Using the distributive property of the cross product:

\( (\vec{a} \times \vec{b}) + \left[ (\vec{b} \times \cos^2 \theta \, \vec{a}) + (\vec{b} \times \sin^2 \theta \, \vec{b}) \right] + \left[ (\cos^2 \theta \, \vec{a} \times \vec{a}) + (\sin^2 \theta \, \vec{b} \times \vec{a}) \right] \).

Since \( \vec{b} \times \vec{b} = 0 \) and \( \vec{a} \times \vec{a} = 0 \), we are left with:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta \, (\vec{b} \times \vec{a}) + \sin^2 \theta \, (- \vec{a} \times \vec{b}) \).

Substitute \( \vec{b} \times \vec{a} = - (\vec{a} \times \vec{b}) \) into the expression:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta \, (- \vec{a} \times \vec{b}) + \sin^2 \theta \, (- \vec{a} \times \vec{b}) \).

Factor out \( \vec{a} \times \vec{b} \):

\( \vec{a} \times \vec{b} \left[ 1 - \cos^2 \theta - \sin^2 \theta \right] \).

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the expression becomes:

\( \vec{a} \times \vec{b} [1 - 1] = 0 \).

∴ The final result is \( \vec{0} \).

Hence, the correct answer is option 1.