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Consider the tetrahedron in the image: The tetrahedron 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.

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    $\begingroup$ The volume of a tetrahedron is not given by "base $\times$ height". That formula only works for constant cross sectional solids. $\endgroup$ – user137731 Jan 7 '16 at 21:07
  • $\begingroup$ @Bye_World, if I wanted to arrive at the given equation from my previous answer of $\frac12|a×b|×|c| \cos \theta $, is it possible? What formula do you suggest I use to find the volume of a tetrahedron? $\endgroup$ – CCC Jan 7 '16 at 21:10
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    $\begingroup$ You can still use $\dfrac{1}{3}Bh$ as the volume of a tetrahedron. $\endgroup$ – user19405892 Jan 7 '16 at 21:10
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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.

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  • $\begingroup$ if I wanted to arrive at the given equation from my previous answer of $\frac12|a×b|×|c| \cos \theta $, is it possible? @user19405892 $\endgroup$ – CCC Jan 7 '16 at 21:07
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    $\begingroup$ Use $V_{\text{tetrahedron}} = \dfrac{1}{3}Bh$. $\endgroup$ – user19405892 Jan 7 '16 at 21:12
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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.

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