Area of triangle in $\mathbb R^3$ given $3$ coordinates I know that if we have $3$ points $a$, $b$,and $c$ in $\mathbb R^3$, the area of the triangle is given by: $\frac{1}{2}\|\vec{ab}\times \vec{ac}\|$. 
This means that the area of the triangle equals half the length of the normal of the plane on which the triangle lays. 
I don't quite understand how this can be. What other geometrical interpretation can be made of $\frac{1}{2}\|\vec{ab}\times \vec{ac}\|$?

 A: The geometric interpretation of the magnitude of the cross product $\left\|\vec v \times \vec w\right\|$ is the area of the parallellogram spanned by the vectors $\vec v$ and $\vec w$ and the triangle with vertices $\vec o$, $\vec v$ and $\vec w$ is exactly half of that parallellogram, hence its area is given by  $\tfrac{1}{2}\left\|\vec v \times \vec w\right\|$.
This interpretation can be clear from the definition (depending on how you define the cross product), or it follows from the formula $\left\|\vec v \times \vec w\right\|=\left\|v\right\|\left\|v\right\|\sin\theta$ ("base times height") where $\theta$ is the angle between $\vec v$ and $\vec w$.
A: Every triangle can be completely specified by two vectors sharing a common vertex. This is because the third side is found by vector subtraction of the other two sides. The triangle spanned by the two vectors is the area we are looking for. By mirrorflipping the two sides across the vector subtraction of the two sides you create a parallelogram. You can see by symmetry that the area of the parallelogram is twice of the area of the triangle. In cartesian coordinates the area of a parallelogram spanned by two vectors with a common vertex is $|ab \times ac|$. Therefore the area of the triangle is $\frac{1}{2}|ab \times ac|$.
