Let a, b, c denote the lengths of the sides of a triangle such that $(a-b)\vec u + (b-c)\vec v + (c-a) (\vec u ×\vec v) =\vec 0$ for any two non-collinear vectors $\vec u$ and $\vec v$, then the triangle is
Answer & explanation
Correct answer: option 2
Since, $\vec u,\vec v$ and $\vec u ×\vec v$ are non-coplanar vectors.
$∴(a-b)\vec u + (b −c)\vec v + (c-a) (\vec u ×\vec v) =\vec 0$
$⇒a-b=0=b-c=c-a$
$⇒a=b=c$
⇒ Triangle is equilateral.