Let \(\vec{a}\) and \(\vec{b}\) be unit vectors inclined at \(30°\). What is the area of the parallelogram whose sides are represented by the vectors \(\vec{a} + 3\vec{b}\) and \(3\vec{a} + \vec{b}\)?
\(4\) square units
The area equals \(|(\vec{a}+3\vec{b})\times(3\vec{a}+\vec{b})|\). Expanding, \((\vec{a}+3\vec{b})\times(3\vec{a}+\vec{b})=(\vec{a}\times\vec{b})-9(\vec{a}\times\vec{b})=-8(\vec{a}\times\vec{b})\), so the area is \(8|\vec{a}\times\vec{b}|=8(1)(1)\sin30°=8\times\dfrac12=4\) square units.
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