cross productdeterminantlinear algebramatrices
This may be a stupid question to some, but when i calculate a cross product of two vectors.
For example the first coordinate of the solution. I put my finger on the first line, then i calculate something that seems like the determinant of a 2×2 matrix.
Is there any connection between matrices and the cross product?
Sorry if that is a stupid qustion, but I am in the second semester and haven't found an answer in the internet.
Thank you for your help!
Perhaps understanding the following definition of the cross product would eliminate your confusion: For two vectors $a$ and $b$ in $\mathbb{R}^3$ the function from $\mathbb{R}^3$ to $\mathbb{R}$ determined by the rule $c \mapsto \det[a, b, c]$ is a linear form on $\mathbb{R}^3$, that is, it is a real-valued linear function on $\mathbb{R}^3$. As such, it can be shown that there is a unique vector in $\mathbb{R}^3$ called the cross product of $a$ and $b$, denoted by $a \times b$, such that $\langle a \times b, c \rangle = \det[a, b, c]$ where $\langle \cdot, \cdot \rangle$ denotes the standard Euclidean inner product on $\mathbb{R}^3$
Not only does this definition elucidate precisely how the cross product is related to the determinant, but determining the "orientation" of the normal vector doesn't depend on the rather bizarre notion of curling your fingers around something to see which way your thumb is pointing. Also, presuming the basic properties of the determinant, many well-known properties of the cross product, which are often proved by various geometric arguments, are immediate.
A little bit more of the 'how and why': the dot product comes about as a natural answer to the question: 'what functions do we have that take two vectors and produce a number?' Keep in mind that we have a natural additive function (vector or componentwise addition) that takes two vectors and produces another vector, and another natural multiplicative function (scalar multiplication) that takes a vector and a number and produces a vector. (We might also want another function that takes two vectors and produces another vector, something more multiplicative than additive — but hold that thought!) For now we'll call this function $D$, and specifically use the notation $D({\bf v},{\bf w})$ for it as a function of the two vectors ${\bf v}$ and ${\bf w}$.
So what kind of properties would we want this hypothetical function to have? Well, it seems natural to start by not distinguishing the two things it's operating on; let's make $D$ symmetric, with $D({\bf v},{\bf w})=D({\bf w},{\bf v})$. Since we have convenient addition and multiplication functions it would be nice if it 'played nice' with them. Specifically, we'd love it to respect our addition for each variable, so that $D({\bf v}_1+{\bf v}_2,{\bf w}) = D({\bf v}_1,{\bf w})+D({\bf v}_2,{\bf w})$ and $D({\bf v},{\bf w}_1+{\bf w}_2) = D({\bf v},{\bf w}_1)+D({\bf v},{\bf w}_2)$; and we'd like it to commute with scalar multiplication similarly, so that $D(a{\bf v}, {\bf w}) = aD({\bf v}, {\bf w})$ and $D({\bf v}, a{\bf w}) = aD({\bf v}, {\bf w})$ — these two conditions together are called linearity (more accurately, 'bilinearity': it's linear in each of its arguments). What's more, we may have some 'natural' basis for our vectors (for instance, 'North/East/up', at least locally), but we'd rather it weren't tied to any particular basis; $D({\bf v},{\bf w})$ shouldn't depend on what basis ${\bf v}$ and ${\bf w}$ are expressed in (it should be rotationally invariant). Furthermore, since any multiple of our function $D$ will satisfy the same equations as $D$ itself, we may as well choose a normalization of $D$. Since $D(a{\bf v},a{\bf v}) = aD({\bf v},a{\bf v}) = a^2D({\bf v},{\bf v})$ it seems that $D$ should have dimensions of (length$^2$), so let's go ahead and set $D({\bf v},{\bf v})$ equal to the squared length of ${\bf v}$, $|{\bf v}|^2$ (or equivalently, set $D({\bf v},{\bf v})$ to $1$ for any unit vector ${\bf v}$; since we chose $D$ to be basis-invariant, any unit vector is as good as any other).
But these properties are enough to define the dot product! Since $$\begin{align} |{\bf v}+{\bf w}|^2 &= D({\bf v}+{\bf w},{\bf v}+{\bf w}) \\ &= D({\bf v}+{\bf w},{\bf v})+D({\bf v}+{\bf w},{\bf w}) \\ &= D({\bf v},{\bf v})+D({\bf w},{\bf v})+D({\bf v},{\bf w})+D({\bf w},{\bf w})\\ &= D({\bf v},{\bf v})+2D({\bf v},{\bf w})+D({\bf w},{\bf w}) \\ &= |{\bf v}|^2+|{\bf w}|^2+2D({\bf v},{\bf w}) \end{align}$$ then we can simply set $D({\bf v},{\bf w}) = {1\over2} \bigl(|{\bf v}+{\bf w}|^2-|{\bf v}|^2-|{\bf w}|^2\bigr)$. A little arithmetic should convince you that this gives the usual formula for the dot product.
While the specific properties for the cross product aren't precisely the same, the core concept is: it's the only function that satisfies a fairly natural set of conditions. But there's one broad catch with the cross-product — two, actually, though they're related. One is that the fact that the cross product takes two vectors and produces a third is an artifact of $3$-dimensional space; in general the operation that the cross-product represents (orthogonality) can be formalized in $n$ dimensions either as a function from $(n-1)$ vectors to a single result or as a function from $2$ vectors that produces a 2-form, essentially a $n(n-1)/2$-dimensional object; coincidentally when $n=3$ this means that the cross-product has the 'vector$\times$vector$\rightarrow$vector' nature that we were looking for. (Note that in $2$ dimensions the natural 'orthogonality' operation is essentially a function from one vector to one vector — it takes the vector $(x,y)$ to the vector $(y,-x)$!) The other catch is lurking in the description of the cross product as a 2-form; it turns out that this isn't quite the same thing as a vector! Instead it's essentially a covector - that is, a linear function from vectors to numbers (note that if you 'curry' the dot-product function $D$ above and consider the function $D_{\bf w}$ such that $D_{\bf w}({\bf v}) = D({\bf v},{\bf w})$, then the resulting object $D_{\bf w}$ is a covector). For most purposes we can treat covectors as just vectors, but not uniformly; the most important consequence of this is one that computer graphics developers have long been familiar with: normals don't transform the same way vectors do! In other words, if we have ${\bf u} = {\bf v}\times{\bf w}$, then for a transform $Q$ it's not (necessarily) the case that the cross product of transformed vectors $(Q{\bf v})\times(Q{\bf w})$ is the transformed result $Q{\bf u}$; instead it's the result ${\bf u}$ transformed by the so-called adjoint of $Q$ (roughly, the inverse of $Q$, with a few caveats). For more background on the details of this, I'd suggest looking into exterior algebra, geometric algebra, and in general the theory of linear forms.
ADDED: Having spent some more time thinking about this over lunch, I think the most natural approach to understanding where the cross product 'comes from' is through the so-called volume form: a function $V({\bf u}, {\bf v}, {\bf w})$ from three vectors to a number that returns the (signed) volume of the rhomboid spanned by ${\bf u}$, ${\bf v}$, and ${\bf w}$. (This is also the determinant of the matrix with ${\bf u}$, ${\bf v}$, and ${\bf w}$ as its columns, but that's a whole different story...) Specifically, there are two key facts:
Now, the fact that the volume form $V({\bf u}, {\bf v}, {\bf w})$ is linear means that we can do the same sort of 'currying' that we talked about above and, for any two vectors ${\bf v}$ and ${\bf w}$, consider the function $C_{\bf vw}$ from vectors ${\bf u}$ to numbers defined by $C_{\bf vw}({\bf u}) = V({\bf u}, {\bf v}, {\bf w})$. Since this is a linear function (because $V$ is linear, by point 2), we know that we have some vector ${\bf c}$ such that $C_{\bf vw} = D_{\bf c}$ (by point 1). And finally, we define the cross product of the two vectors ${\bf v}$ and ${\bf w}$ as this 'vector' ${\bf c}$. This explains why the cross product is linear in both of its arguments (because the volume form $V$ was linear in all three of its arguments) and it explains why ${\bf u}\times{\bf v} = -{\bf v}\times{\bf u}$ (because $V$ changes sign on swapping two parameters). It also explains why the cross product isn't exactly a vector: instead it's really the linear function $C_{\bf vw}$ disguising itself as a vector (by the one-to-one correspondence through $D_{\bf c}$). I hope this helps explain things better!
Best Answer
Maybe this isn't the answer you're looking for, but one expression for the determinant of a 3x3 matrix with columns $\vec v_1,\vec v_2,\vec v_3$ is $$ \vec v_1\cdot(\vec v_2\times\vec v_3) $$ You can make sense of this algebraically or geometrically (recall that the determinant is the volume of a parallelipiped whose sides are given by the three vectors).