# Trigonometry - Calculating the pyramid volume

The problem:

There be the points $P_0(0,0,0)$, $P_1(1,1,1)$, $P_2(2,-1,2)$ and $P_3(3,0,1)$. Calculate the volume of the pyramid.

Now I assumed the base of the pyramid is a triangle, with points $P_1$, $P_2$ and $P_3$.

So I know $\underline u:=\overrightarrow{P_1P_2} = <1, -2, 1>$ and $\underline v:=\overrightarrow{P_1P_3}=<2,-1,0>$.

I calculated the angle between $\underline u$ and $\underline v$:

$$\cos\theta:=\frac{\underline{u}\cdot\underline{v}}{||\underline{u}||\,||\underline{v}||}=\frac{4}{\sqrt{30}}\Longrightarrow \theta=43.0887^\circ$$ $$S_\Delta=\frac{||\underline{u}||\,||\underline{v}||\sin\theta}{2}=1.87$$

and I thought the pyramid's height to be $\,2\,$ , so the volume is $\,\,\displaystyle{V=\frac{S_\Delta\cdot 2}{3}=1.246}$

I believe I'm somewhat wrong with this.

Help will be much appreciated!

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This problem is relatively simple because one of your points is at the origin. The volume of your pyramid is then equal to $1/6$ of the determinant of the matrix formed by the other points:
$$V = \frac{1}{6} \left |\begin{array}\\1&1&1\\2&-1&2\\3&0&1 \end{array} \right |$$
Evaluating the determinant, I get $V=1$.