Let $\vec{OA}=\vec{a}, \vec{OB}=\vec{b}$ and $\vec{OC}=\vec{c}$ and $|\vec{a}|=3, |\vec{b}|=2$ and $|\vec{c}|=4$. We also know that $\measuredangle(\vec{a},\vec{b})=\measuredangle(\vec{a},\vec{c})=\measuredangle(\vec{b},\vec{c})=\dfrac{\pi}{3}.$ Given the affine coordinate system $O{\vec{a}\vec{b}\vec{c}},$ find the coordinates of the vector $\vec{a}\times\vec{b}$.
We could easily find that $\vec{a}\cdot\vec{b}=3, \vec{a}\cdot\vec{c}=6$ and $\vec{b}\cdot\vec{c}=4$.
Also the Gram determinant gives the square of the triple product of the vectors which I found to be $[\vec{a}\;\vec{b}\;\vec{c}]^2=9\cdot16\cdot2.$
Let $\vec{d}=\vec{a}\times\vec{b}$. Then, by definition, we have $\vec{d}\perp\vec{a}$ and $\vec{d}\perp\vec{b}$. Let $\vec{d}=\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}$. Multiplying the last equality by $\vec{a}$ gives $$3\alpha+\beta+2\gamma=0$$ In the same way by multiplying by $\vec{b}$ we get $$3\alpha+4\beta+4\gamma=0$$ What else can we do though (we need one more equation)?
Can we multiply an equation by a vector product? Or maybe we could try to multiply by $\vec{c}$ to get $$(\vec{a}\times\vec{b})\vec{c}=(\vec{a}\vec{b}\vec{c})=\pm12\sqrt{2}=6\alpha+4\beta+16\gamma$$ Am I right? How do I determine the exact sign of the triple product, though?