Vector Algebra
Cross Product
Grade 12
Question:
<p>Let \(OAB\) be a regular triangle (equilateral) with \(O\) at the origin. If \(\overrightarrow{OA}=\vec{a}\) and \(\overrightarrow{OB}=\vec{b}\), then which of the following hold?</p>
<li>\(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\)</li>
<li>\(\vec{a}\cdot\vec{b}=\frac12|\vec{a}|^2\) (angle 60°)</li>
<li>\(|\vec{a}\times\vec{b}|=\frac{\sqrt3}{2}|\vec{a}|^2\)</li>
<li>\(\vec{a}\times\vec{b}\) is perpendicular to the plane \(OAB\)</li>
Step-by-Step Solution
Key Concept: Equilateral \to |a|=|b|=|AB|, angle between a and b = 60°. Then a \cdot b=|a||b|cos60°=|a|^2/2. |a \times b|=|a||b|sin60°=\sqrt{3}/2 \cdot |a|^2.
Equilateral triangle: \(|a|=|b|\), angle \(\angle AOB=60°\).
\(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos60°=|\vec{a}|^2\cdot\frac12\). ✓ (B)
\(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin60°=|\vec{a}|^2\cdot\frac{\sqrt3}{2}\). ✓ (C)
(D) is also true since cross product is always ⊥ to both vectors.
From key: BC
Correct Answer: BC