Volume of tetrahedron using cross and dot product 
Consider the tetrahedron in the image:  Prove that the volume of the tetrahedron is given by $\frac16 |a \times b \cdot c|$.

I know volume of the tetrahedron is equal to the base area times height, and here, the height is $h$, and I’m considering the base area to be the area of the triangle $BCD$. 
So, what I have is:
$$\begin{align}
\text{base area} &= \frac12 \lvert a \times b \rvert \\
\text{height $h$} &= \lvert c\rvert \cos \theta
\end{align}$$
So volume is $$V=\frac12 \lvert a \times b\rvert \cdot \lvert c\rvert \cos \theta $$
But I don’t know how to arrive from this at $\frac16 |a \times b \cdot c|$.
Please advise.
 A: Hint: $\mathbf{a}\times\mathbf{b}$ "points straight up".  It is therefore parallel to the line $h$ that you show.  Therefore, it has the same angle with $c$ as $h$ does.
A: Here is one way to think of it. A tetrahedron is $\dfrac{1}{6}$ of the volume of the parallelipiped formed by $\vec{a},\vec{b},\vec{c}$. The volume of the parallelepiped is the scalar triple product $|(a \times b) \cdot c|$. Thus, the volume of a tetrahedron is $\dfrac{1}{6} |(a \times b) \cdot c|$
In order to solve the question like you are trying to, notice that by $V = \dfrac{1}{3}Bh = \dfrac{1}{6}||a \times b|| \cdot h$. Then, $h = ||c|| \cdot |\cos(\theta)|$. Thus, we have $V = \dfrac{1}{6}||a \times b|| \cdot ||c|| \cdot |\cos(\theta)|$. Now see that $|c \cdot (a \times b)| = ||c|| \cdot ||(a \times b)|| \cdot |\cos(\theta)|$ and thus $V = \dfrac{1}{6}|(a \times b) \cdot c|$.
I'd also like to say that the notation you are using is a little weird. In order to avoid confusion, $|x|$ denotes absolute value and $||x||$ denotes magnitude.
A: The volume will be (area of the base) x 1/3 height ----------------(1)
now in this case if we take triangle BCD as base its area will be
= 1/2 |a||b||sin(m)| = 1/2 |a x b| ---------------------(2)
where m is the angle between a and b therefore b sin(m) will be the length of the projection of b perpendicular to a ,i.e., height of the triangle BCD if we take a as its base.
now the height of the tetrahedron will be the projection of c perpendicular to the base
unit vector that is perpendicular to base BCD is = (a x b)/|a x b| = n
therefore height of the tetrahedron(h) which can be expressed as projection of c on n
h = |c||n||cos(θ)| = c.n = c.(a x b)/|a x b| -------------(3)
from eq 1,2 and 3 we have
volume = 1/2|a x b| 1/3[c.(a x b)]/|a x b|
= 1/6[c.(a x b)]
= 1/6[(a x b).c]
PS- I'm still in high school and it's my first post here so I'm sorry if I messed up.
