the volume of the tetrahedron How can I prove the following statement?

Prove that the volume of the tetrahedron formed by four non-coplanar points $A_i=(x_i,y_i,z_i),$ $1 \le i \le 4$, is equal to $\frac{1}{6} \left| \overrightarrow{A_1 A_2} \times \overrightarrow{A_1 A_3} .\overrightarrow{A_1 A_4}\right|.$

we did volume of the paralelogram at the before lesson and this is our homework but I have no idea where to start.
 A: We know that the volume of a tetrahedron is $V={1\over3}A_bh$ where $A_b$ is the area of the base and $h$ is the height from the base to the opposite vertex. If $\overrightarrow{A_1}$, $\overrightarrow{A_2}$ and $\overrightarrow{A_3}$ are the vectors pointing the vertex of the base, then $\overrightarrow{A_1 A_2}=\vec a$, $\overrightarrow{A_1 A_3}=\vec b$ and $\overrightarrow{A_2 A_3}=\vec c$ are the vectors along the base sides with mignitude $a$, $b$, $c$ respectively; so you can write the area of the base (it is a triangle) as:
$$A_b={1\over2}a b\sin\gamma=\left({1\over2}a c\sin\beta={1\over2}bc\sin\alpha\right)$$
so we have $A_b={1\over2}|\vec a\times \vec b|$.
The vector $\vec {A_b}$ is normal to the base and thus parallel to $h$; hence we can project $\overrightarrow{A_1 A_4}=\vec d$ (the vector from the vertex "1" to the apex of the tetrahedron) along $\vec {A_b}$ to obtain $h$:
$$h=\vec d|\cos\delta|$$
where $\delta$ is the angle between $\overrightarrow{A_1 A_4}=\vec d$ and the height (or we can say, the direction of $\vec {A_b}$). I put $|\cos\delta|$ because $\vec {A_b}$ might be upwards or downwards, so the angle could be $\delta$ or $\pi-\delta$ and $|\cos\delta|=|\cos(\pi-\delta)|$.
The volume will be:
$$V={1\over3}{1\over2}|\vec a\times \vec b|d|\cos\delta|={1\over6}d|\vec a\times \vec b||\cos\delta|={1\over6}|\vec d\cdot (\vec a\times \vec b)|={1\over6}|\overrightarrow{A_1 A_4}\cdot (\overrightarrow{A_1 A_2}\times \overrightarrow{A_1 A_3})|$$
