<p>Let \(a, b, c\) denote the lengths of the sides of a triangle such that \((a-b)\vec{u} + (b-c)\vec{v} + (c-a)(\vec{u} \times \vec{v}) = 0\) for any two non-collinear vectors \(\vec{u}\) and \(\vec{v}\), then the triangle is</p>
Step-by-Step Solution
Key Concept: Non-coplanar vectors are linearly independent, so coefficients in their linear combination must all be zero for the equation to hold universally.
Solution: Since \(\vec{u}, \vec{v}\) and \(\vec{u} \times \vec{v}\) are non-coplanar vectors, they form a linearly independent set. For the equation \((a-b)\vec{u} + (b-c)\vec{v} + (c-a)(\vec{u} \times \vec{v}) = 0\) to hold for any two non-collinear vectors \(\vec{u}\) and \(\vec{v}\), we must have: \(a - b = 0\), \(b - c = 0\), and \(c - a = 0\) Therefore: \(a = b = c\) So, the triangle is equilateral. ∴ Answer is (b).
Correct Answer: B