I have designed three vectors that consist of:
\begin{align}
a &= -4i-10j-k\\
b &= 7i+9j-2k\\
c &= 3i+9j+4k\\
\end{align}
And the task was to find the volume that was created from the above vectors when creating a parallelepiped. The current method I have used is modulus matrix of the scalar triple product ($a$.$b*c$) and also the distributive law. the results concluded from this was $88$ units cubed, but I was wondering whether there were any other methods I could use to verify my answer?
[Math] Volume of a Parallelepiped, Vectors
vectorsvolume
Related Solutions
Let us denote by $\vec{A},\vec{B},\vec{C}$ the vectors issued from the common vertex, with resp. length (=norms) $a,b,c$.
Let us define matrix
$$M=[\vec{A}|\vec{B}|\vec{C}]$$
(in the RHS, columns entries are the coordinates of $\vec{A},\vec{B},\vec{C}$ with respect to a certain orthonormal basis).
It is well known that $\det(M)=V$ (parallelepiped's volume).
Now compute (in close connection with the computations of paragraph "volume" in this reference) the so-called "Gram matrix" of ordered system $(\vec{A},\vec{B},\vec{C})$:
$$M^T M=\begin{pmatrix}\vec{A}\cdot\vec{A}&\vec{A}\cdot\vec{B}&\vec{A}\cdot\vec{C}\\ \vec{B}\cdot\vec{A}&\vec{B}\cdot\vec{B}&\vec{B}\cdot\vec{C}\\ \vec{C}\cdot\vec{A}&\vec{C}\cdot\vec{B}&\vec{C}\cdot\vec{C} \end{pmatrix}$$
where the dots are for dot product.
Using the given angles :
$$M^T M=\begin{pmatrix}a^2&ab \cos \beta&ac \cos \alpha\\ ab \cos \beta&b^2&bc \cos \alpha\\ ac \cos \alpha&bc \cos \alpha&c^2 \end{pmatrix}$$
Equating the determinants of the LHS and the RHS (remember that $\det(M^TM)=\det(M)^2$) :
$$V^2=a^2b^2c^2(1+2 \cos^2 \alpha \cos \beta-2\cos^2 \alpha-\cos^2 \beta).$$
Remark : Condition $\alpha > 2\beta$ hasn't been used.
The method at that site ignores the sign of the determinant, and is phrased in a way that makes the generalization to higher dimensions a bit less clear than I'd like. I'll try here to provide an altered version of the approach up to four dimensions. I'll use bold for vectors, subscripts for components, and double vertical lines for length, so that $\left\Vert \mathbf{a}\right\Vert =\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots}$.
As was mentioned by LutzL in a comment, this method is very closely connected to using the $QR$-decomposition to find the absolute value of the determinant, as described on Wikipedia here.
1D:
Let's calculate $\det\left(\mathbf{a}\right)$ where $\mathbf{a}$ has one nonzero component. It's $\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a positive component and $-\left\Vert \mathbf{a}\right\Vert$ if $\mathbf{a}$ has a negative component.
2D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b}\right)$ where $\mathbf{a}$ and $\mathbf{b}$ are not collinear.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to $\mathbf{b}$. We set $\mathbf{n}\bullet\mathbf{b}$ equal to $\boldsymbol{0}$. That's two unknowns and only one equation. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ can be chosen freely.)
Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left\Vert \mathbf{b}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is $\pm\mathbf{n}'\bullet\mathbf{a}$ where the $\pm$ sign here (which may not be the sign of the determinant) is positive exactly when the rotation to get from $\mathbf{a}$ to $\mathbf{b}$ is in the same direction (clockwise or counterclockwise) as the rotation to get from $\mathbf{b}$ to $\mathbf{n}'$. Unfortunately, that can't be determined by a dot-product calculation.
3D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c}$ are not coplanar.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to both $\mathbf{b}$ and $\mathbf{c}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c}$ equal to $\boldsymbol{0}$. That's three unknowns and only two equations. In a typical case, the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but then $n_{2}$ or $n_{3}$ can be chosen freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ (this choice differs from the original author), but lies in the same plane as $\mathbf{b}$ and $\mathbf{c}$. To keep it in that plane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c}=\boldsymbol{0}$. Again that's three unknowns and two equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left\Vert \mathbf{c}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{b}$ and $\mathbf{c}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I'm pretty sure the $\pm$ sign is positive when the handedness (right or left) of $\mathbf{a},\mathbf{b},\mathbf{c}$ is the same as that of $\mathbf{c},\mathbf{n}',\mathbf{o}'$.
4D:
Let's calculate $\det\left(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right)$ where $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ are not in the same $3$-dimensional hyperplane.
Let's ignore $\mathbf{a}$ for now. The first step is to find a vector $\mathbf{n}$ that's orthogonal to all three of $\mathbf{b},\mathbf{c},\mathbf{d}$. We set $\mathbf{n}\bullet\mathbf{b},\mathbf{n}\bullet\mathbf{c},\mathbf{n}\bullet\mathbf{c}$ all to $\boldsymbol{0}$. That's four unknowns and only three equations. In a typical case the component $n_{1}$ of $\mathbf{n}$ is not forced to be $0$, so it can be whatever we want that's nonzero (e.g. $1$). (In a special case, $n_{1}$ might be forced to be $0$, but there will be at least one component we could choose freely.)
The second step is to find a vector $\mathbf{o}$ that's orthogonal to $\mathbf{c}$ and $\mathbf{d}$, but lies in the same $3$-dimensional hyperplane as $\mathbf{b},\mathbf{c},\mathbf{d}$. To keep it in that hyperplane, we need $\mathbf{o}$ to be orthogonal to $\mathbf{n}$. So we have $\mathbf{o}\bullet\mathbf{n}=\boldsymbol{0}$ as well as $\mathbf{o}\bullet\mathbf{c},\mathbf{o}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
The third step is to find a vector $\mathbf{p}$ that's orthogonal to $\mathbf{d}$, but lies in the same $2$-dimensional plane as $\mathbf{c},\mathbf{d}$. To keep it in that plane, it should be orthogonal to $\mathbf{n}$ as well as $\mathbf{o}$. So we have $\mathbf{p}\bullet\mathbf{n},\mathbf{p}\bullet\mathbf{o}=\boldsymbol{0}$ as well as $\mathbf{p}\bullet\mathbf{d}=\boldsymbol{0}$. Again that's four unknowns and three equations, so we have a degree of freedom and could choose a particular value for some component.
Now scale $\mathbf{p}$ to get a new vector $\mathbf{p}'$ so that $\left\Vert \mathbf{p}'\right\Vert =\left\Vert \mathbf{d}\right\Vert$ . By some geometry, the area of the parallelogram formed by $\mathbf{c}$ and $\mathbf{d}$ is then $\left|\mathbf{p}'\bullet\mathbf{c}\right|$. Now scale $\mathbf{o}$ to get a new vector $\mathbf{o}'$ so that $\left\Vert \mathbf{o}'\right\Vert =\left|\mathbf{p}'\bullet\mathbf{c}\right|$. By some geometry, the volume of the parallelepiped formed by $\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{o}'\bullet\mathbf{b}\right|$. Now scale $\mathbf{n}$ to get a new vector $\mathbf{n}'$ so that $\left\Vert \mathbf{n}'\right\Vert =\left|\mathbf{o}'\bullet\mathbf{b}\right|$. By some geometry, the hypervolume of the hyperparallelepiped formed by $\mathbf{a},\mathbf{b}, \mathbf{c}, \mathbf{d}$ is then $\left|\mathbf{n}'\bullet\mathbf{a}\right|$. The determinant is then $\pm\mathbf{n}'\bullet\mathbf{a}$ where I think the $\pm$ sign is positive when the orientation of $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ is the same as that of $\mathbf{d},\mathbf{n}',\mathbf{o}',\mathbf{p}'$.
Best Answer
$$V=|-4\cdot9\cdot4+(-1)\cdot7\cdot9+3\cdot(-10)\cdot(-2)-3\cdot9\cdot(-1)-(-4)\cdot9\cdot(-2)-4\cdot7\cdot(-10)|=88$$ It's the best way, I think.
We can also just to prove this formula.
Id est, let $ABCDA'B'C'D'$ be our parallelepiped.
We can find $S_{ABCD}=AB\cdot AC\cdot\sin\measuredangle BAC$ and calculate an altitude $A'K$ to $ABCD$ of the parallelepiped.
By this way we calculate a module of the scalar triple product of our vectors.