Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be unit vectors such that \(\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}=0\). If the angle between \(\vec{b}\) and \(\vec{c}\) is \(\pi/6\), then what is \(\vec{a}\) equal to?
\(2(\vec{b}\times\vec{c})\)
Since \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), it must be parallel to \(\vec{b}\times\vec{c}\). As \(\vec{b}\) and \(\vec{c}\) are unit vectors with angle \(\pi/6\) between them, \(|\vec{b}\times\vec{c}|=\sin(\pi/6)=1/2\). Since \(\vec{a}\) is a unit vector, \(\vec{a}=\dfrac{\vec{b}\times\vec{c}}{|\vec{b}\times\vec{c}|}=2(\vec{b}\times\vec{c})\).
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