Finding the volume of a pyramid (the vector way)

Question

The problem

I have 4 points

$ P \; (-1,2,0) \\ Q \; (2,1,3) \\ R \; (1,0,1) \\ S \; (3,-2,3) $

and I want to find the volume of a pyramid. What I'm most concerned here is the appropriate strategy to go about this.

What I've tried

1. Since the volume is calculated by $\mbox{base} \cdot \mbox{height} \times \frac{1}{3}$ I first find the base
2. The base in this case $\frac{1}{2}\lVert PQ\times QR \rVert$
3. Next the height vector will co collinear with the normal vector, so I find that from the cross product ${\bf n} = PQ\times QR$

By this point, I've decided that $S$ is the top of the pyramid, and $PQR$ is the base. What I struggle with is finding the distance between the top and base, I.e. the height.

In my mind, that has to be a scaled version of the $\bf n$, but how do I find the scale? Or is my approach broken?

Answer 1

Hint: The volume of a Parallelepiped is given by the determinant of three of its spanning vectors. (That is $Q-P$, $R-P$ and $S-P$, for instance)

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 'Move' $S$ to the origin of coordinates. We get the points $\vec{p} = \pars{-2,4,-3}$, $\vec{q} = \pars{-1,3,0}$, $\vec{r} = \pars{-2,2,-2}$ and $\vec{s} = \pars{0,0,0}$ which correspond to the "old" points $P$, $Q$, $R$ y $S$, respectively. With the 'new' points: $$ V = \int_{V}\dd V = \int_{V}{\nabla\cdot\vec{r} \over 3}\,\dd V = {1 \over 3}\int_{S}\vec{r}\cdot\dd\vec{S}\tag{1} $$ where we used Gauss's theorem. The integration over the 'walls' vanishes out since $\vec{r}\cdot\dd\vec{S} = 0$. There remains an integration over a triangle $\pars{~\mbox{the pyramid base}\ B~}$which vertexes at $\vec{p}$, $\vec{q}$ and $\vec{r}$.

$\pars{1}$ is reduced to $\ds{V = {1 \over 3}\int_{B}\vec{r}\cdot\hat{n}\,\dd S}$ where $\hat{n}$ is a perpendicular unit vector to $B$. $\pars{~\dd S \equiv \verts{\dd\vec{S}}~}$:

$$ \hat{n} \equiv {\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}} \over \verts{\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}}}\,,\quad \left\vert% \begin{array}{rcl} \pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}} & = & \pars{-1,1,-3}\times\pars{0,2,-1} \\ & = & \pars{5,-1,-2} \\[2mm] \verts{\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}} & = & \root{30} \\[2mm] \hat{n} & = & {1 \over \root{30}}\,\pars{5,-1,-2} \end{array}\right. $$ Then, $$ V = {\root{30} \over 90}\int_{B}\pars{5x - y -2z}\dd S $$ Can you take from here ?