Question:
<p>In an isosceles triangle \(ABC\) with \(AB=AC\), M is the midpoint of \(BC\). Given an equilateral triangle with side \(\vec{a}\), the scalar triple product \([\vec{a},\vec{b},\vec{c}]\) (where \(\vec{a},\vec{b},\vec{c}\) are sides) equals</p>
\(0\)
\(a^3\sqrt2\)
\(a^3/2\)
\(a^3\sqrt3/2\)
Step-by-Step Solution
Key Concept: The scalar triple product [a,b,c] with a+b+c=0 (closed triangle) is zero since the vectors are coplanar.
For any triangle, the three side vectors satisfy $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ (closed polygon). Coplanar vectors have scalar triple product = 0.
$[\vec{a},\vec{b},\vec{c}]=\vec{a}\cdot(\vec{b}\times\vec{c})=0$ (coplanar). Answer: (A)
Correct Answer: A