determination of the volume of a parallelepiped Here is a  parallelepiped.I want to determine the volume of the parallelepiped.

One of my friends said to me that the volume of the parallelepiped  can be found out by the following formula.

$${\rm Volume}=|\vec a\cdot(\vec b\times \vec c)|$$

I can not understand  why and how this formula works? Can anyone explain with clarification.
 A: Think of it this way: Area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is given by $|\vec{b} \times \vec{c}|$. This is easy to see because the area of a parallelogram is $base \times height = c b \sin \theta $ where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$. 
In order to get the volume, we need to multiply this area by the projection of $\vec{a}$ along the direction perpendicular to the parallelogram formed by $\vec{b}$ and $\vec{c}$ which is given by the formula you have mentioned because $ \vec{b} \times \vec{c} $ is a vector perpendicular to the plane (the idea is the same - $base \times height$ - the base is now an area rather than a length)
A: $$ e_i\times e_{i+1}:=e_{i+2}$$
So we have $$ b\times c=|b||c|\sin\ \theta \overrightarrow{n}$$
where $\theta$ is an angle between $b$ and $c$ and $\overrightarrow{n}$ is an unit normal to the plane $P$ of $b$ and $c$ 
Here $|b||c|\sin\ \theta$ is area of parallelpipe of $b$ and $c$ 
And $|a\cdot \overrightarrow{n}|$ is the height by $a$ in normal direction to $P$. 
A: $\newcommand{\Reals}{\mathbf{R}}\DeclareMathOperator{\Vol}{Vol}$Here's a sketch of a typical "modern" argument that generalizes to parallelipipeds in $\Reals^{n}$ for arbitrary positive integers $n$.
The idea is to introduce two functions of ordered triples of vectors in $\Reals^{3}$:


*

*The "signed volume" $\Vol(a, b, c)$ of the parallelipiped spanned by the vectors (in the given order);

*The "triple product"
$$
a \cdot (b \times c) = \det\left[\begin{matrix}
  a_{1} & a_{2} & a_{3} \\
  b_{1} & b_{2} & b_{3} \\
  c_{1} & c_{2} & c_{3} \\
  \end{matrix}\right];
$$
and to show these functions are identical. (The answer to the original question follows at once by taking absolute values.)
Guided by retrospect, let's consider a general real-valued function $f$ of three vectors. We say $f$ is:


*

*Multilinear if "$f$ is linear in each variable separately"; that is,
$$
f(ta_{1} + a_{2}, b, c) = tf(a_{1}, b, c) + f(a_{2}, b, c)\quad
\text{for all $a_{1}$, $a_{2}$, $b$, $c$ in $\Reals^{3}$, all real $t$,}
$$
and similarly in the second and third variables.

*Skew-symmetric if exchanging a pair of arguments changes the overall sign; that is,
$$
f(a, c, b) = f(c, b, a) = f(a, c, b) = -f(a, b, c),\quad
\text{for all $a$, $b$, $c$ in $\Reals^{3}$;}
$$

*Normalized if $f$ evaluates to unity on the standard basis: $f(e_{1}, e_{2}, e_{3}) = 1$.
Notes:


*

*Signed volume is a geometric measure of "oriented volume" spanned by an ordered triple of vectors in $\Reals^{3}$ (or, more generally, by an ordered $n$-tuple of vectors in $\Reals^{n}$). We define signed volume to be multilinear, skew-symmetric, and normalized. Normalization says the signed volume of the unit cube (with spanning vectors in standard order) is unity. Skew-symmetry says that exchanging the order of two vectors changes the sign of the volume, like mirror reflection. Multilinearity contains most of the interesting geometry: If we fix all but one of the vectors spanning a parallelipiped and examine how volume depends on the remaining vector, then (i) Multiplying the free vector by a scalar $t$ multiplies the signed volume by $t$; (ii) Shearing the free vector in directions parallel to the fixed edges does not change the signed volume.

*These properties contain some redundancies; for example, linearity in the first variable together with skew-symmetry implies linearity in the second and third variables.

*An ordered triple $(a, b, c)$ of vectors can be put into $3! = 6$ orderings. If the reordering is effected by an even permutation, the sign of $f$ is invariant; if the reordering is effected by an odd permutation, the sign of $f$ changes. In particular, $f(a, a, b) = -f(a, a, b)$ (by swapping the first two arguments), so $f(a, a, b) = 0$ for all $a$ and $b$. Generally, if any two of $a$, $b$, $c$ are equal, then $f(a, b, c) = 0$.

*A multilinear function as above is called a $3$-tensor on $\Reals^{3}$. A non-zero skew-symmetric $3$-tensor on $\Reals^{3}$ is a volume element. (Often, physicists and mathematicians speak of "tensor fields" and "volume forms"; these are functions whose value at each point is, respectively, a tensor or a volume element. To add to the confusion, most people use "tensor" to refer to a "tensor field" when context permits. It's safe to ignore all this if it doesn't make sense; just trying to forestall terminological nitpicking.)
Theorem: There exists a unique normalized, skew-symmetric, multilinear function on ordered triples of vectors in $\Reals^{3}$.
Proof (sketch): Write $a = a_{1}e_{1} + a_{2}e_{2} + a_{3}e_{3}$ as a linear combination in the standard basis and evaluate, using multilinearity to expand:
\begin{align*}
f(a, b, c)
  &= f(a_{1}e_{1} + a_{2}e_{2} + a_{3}e_{3}, b, c) \\
  &= a_{1} f(e_{1}, b, c) + a_{2} f(e_{2}, b, c) + a_{3} f(e_{3}, b, c).
\end{align*}
Continue similarly by substituting $b = b_{1}e_{1} + b_{2}e_{2} + b_{3}e_{3}$ and $c = c_{1}e_{1} + c_{2}e_{2} + c_{3}e_{3}$.
Expanding completely results in $3^{3} = 27$ terms, namely the terms $a_{i} b_{j} c_{k}\, f(e_{i}, e_{j}, e_{k})$ with $(i, j, k)$ an arbitrary ordered triple of indices from $\{1, 2, 3\}$.
Thanks to skew-symmetry, any term with a repeated index is automatically zero. There are precisely $3! = 6$ terms with non-repeated indices: $a_{1} b_{2} c_{3}\, f(e_{1}, e_{2}, e_{3})$, and $a_{2} b_{1} c_{3}\, f(e_{2}, e_{1}, e_{3})$, and four others. By skew-symmetry and normalization,
\begin{align*}
 1 &= f(e_{1}, e_{2}, e_{3}) = f(e_{2}, e_{3}, e_{1}) = f(e_{3}, e_{1}, e_{2}), \\
-1 &= f(e_{1}, e_{3}, e_{2}) = f(e_{3}, e_{2}, e_{1}) = f(e_{2}, e_{1}, e_{3}).
\end{align*}
Our expansion of $f(a, b, c)$ started with $3^{3} = 27$ terms, of which only $3! = 6$ were not eliminated by skew-symmetry. The preceding evaluations of $f$ on permutations of the standard basis imply that
\begin{align*}
f(a, b, c)
  &= a_{1}b_{2}c_{3} - a_{1}b_{3}c_{2}
   + a_{2}b_{3}c_{1} - a_{2}b_{1}c_{3}
   + a_{3}b_{1}c_{2} - a_{3}b_{2}c_{1} \\
  &= a_{1}(b_{2}c_{3} - b_{3}c_{2})
   + a_{2}(b_{3}c_{1} - b_{1}c_{3})
   + a_{3}(b_{1}c_{2} - b_{2}c_{1}) \\
  &= a \cdot (b \times c).
\tag{1}
\end{align*}
The preceding formula says there is at most one normalized, skew-symmetric, multilinear function on triples of vectors in $\Reals^{3}$ (because the formula was derived by assuming these three properties).
Conversely, this formula can be checked to be multilinear (i.e., it satisfies a generalized distributive law), skew-symmetric (swapping two arguments changes the overall sign), and normalized. This completes the proof of the theorem.
Actually, we're almost done:
Corollary: For all $a$, $b$, $c$ in $\Reals^{3}$, $\Vol(a, b, c) = a \cdot (b \times c)$.
Proof (sketch): It suffices to show that both signed volume and the triple product are multilinear, skew-symmetric, and normalized. But as in note 1. above, signed volume is multilinear, skew-symmetric, and normalized, while the work is already done for the triple product, by (1).
Exercise: Work out the details of the case $n = 2$. (Here there are $2^{2} = 4$ terms in the complete expansion of $f(a, b)$, of which $2! = 2$ "survive" elimination by skew-symmetry.) Sketch a general parallelogram in $\Reals^{2}$, and convince yourself geometrically that signed area is multilinear (compare note 1. above).
A: whilst the 3-d vector formulation is very handy for real-world calculations in Euclidean 3-space it is also worth noticing that:
in  1 dimension
$$
V_1(a) =a_1 =|a|
$$
in 2 dimensions
$$
V_2(a,b) = a_1 b_2 - a_2b_1 = |a b|
$$
in 3 dimensions
$$
V_3(a,b,c) = a_1(b_2c_3-b_3c_2) +a_2(b_3c_1-b_1c_3) + a_3(b_1c_2-b_2c_1)=|a b c|
$$
this shows generally how a higher-dimensional volume element is defined in terms of elements in one less dimension, and the formalism can easily be extended to 4 and more dimensions. in this 3-d case it is easy to see that $V_3(a,b,c) = a.b\times c = b.c \times a = c.a \times b$
in fact we may write, using the Levi-Civita epsilon:
$$
V = \sum_{i,j,k} \epsilon_{ijk} \frac{\partial^3 V}{\partial a_i \partial b_j \partial c_k}
$$
interpreting the volume as the determinant of a linear transformation mapping orthonormal basis elements $e_i$ to the edge-vectors of a parallelopiped naturally embodies the multilinear dependence of volume on the edge-vectors as well as the fact that the scalar volume has a sign which depends on orientation. it also relates directly to the use of the Jacobian as the scale factor which must be incorporated when applying a differentiable transform of co-ordinates to evaluate an integral.
