geometryvectors
The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:
$$\left| \begin{matrix}
\vec{i} & \vec{j} & \vec{k} \\
a_i & a_j & a_k \\
b_i & b_j & b_k
\end{matrix}\right|$$
Is there any physical significance to this matrix, or is it just some mathematical trick?
I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?
Perhaps you would find the geometric interpretations of the dot and cross products more intuitive:
The dot product of A and B is the length of the projection of A onto B multiplied by the length of B (or the other way around--it's commutative).
The magnitude of the cross product is the area of the parallelogram with two sides A and B. The orientation of the cross product is orthogonal to the plane containing this parallelogram.
Why can't vectors be divided?
How would you define the inverse of a vector such that $\mathbf{v} \times \mathbf{v}^{-1} = \mathbf{1}$? What would be the "identity vector" $\mathbf{1}$?
In fact, the answer is sometimes you can. In particular, in two dimensions, you can make a correspondence between vectors and complex numbers, where the real and imaginary parts of the complex number give the (x,y) coordinates of the vector. Division is well-defined for the complex numbers.
The cross-product only exists in 3D.
Division is defined in some higher-dimensional spaces too (such as the quaternions), but only if you give up commutativity and/or associativity.
Here's an illustration of the geometric meanings of dot and cross product, from the wikipedia article for dot product and wikipedia article for cross product:
Any equations you set up will be uglier than the cross product (they will basically by expansions of $\times$)
Remember, vectors only can define a direction. The issue with the magnetic field is that it is heavily related to planes, and is a perpendicular force. We can't assign a magnetic field analogous to the electric field (as in: a field which shows the direction of force), as it depends upon the direction of velocity.
The cross product enables us to define a plane with a vector perpendicular to it. Aside from that, it gives perpendicular stuff. This clearly points to the use of the cross product.
Besides, the cross product isn't that dirty. It makes stuff simpler, if anything.
If you want to do it all via equations, use this:
$$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \\ \end{vmatrix}$$ These rules may help you avoid the determinant form for simpler cross products: $$\hat{i}\times\hat{j}=\hat{k};\:\hat{j}\times\hat{k}=\hat{i};\:\hat{k}\times\hat{i}=\hat{j};\:\vec{a}\times\vec{a}=0;\:\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$
Over here, the right hand rule is implied by our choice of a right handed system of $\hat{i} , \hat{j} , \hat{k}$ or $x,y,z$. So you don't need to touch it, it is already inherent in the coordinate system.
Best Answer
The physical (geometric) relevance to the matrix
$$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{matrix}\right|$$
with regard to the cross product $\vec{a} \times \vec{b}$ is
1:that the three vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ constitute a vector basis that spans a space which iseither also spanned by $\vec{a}$, $\vec{b}$, and one additional vector which is perpendicular to $\vec{a}$ as well as $\vec{b}$;
or, in case that vectors $\vec{a}$ and $\vec{b}$ are parallel to each other, also spanned by $\vec{a}$ and two additional (non-parallel) vectors.
2:the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ are pairwise orthogonal (perpendicular) to each other.Therefore vectors $\vec{a}$ and $\vec{b}$ as well as the cross product vector $\vec{a} \times \vec{b}$ can be completely and uniquely expressed in terms of the corresponding components:
$\vec{a} := a_i \vec{i} + a_j \vec{j} + a_k \vec{k}$,
$\vec{b} := b_i \vec{i} + b_j \vec{j} + b_k \vec{k}$, and
$\vec{a} \times \vec{b} := \{ab\}_i \vec{i} + \{ab\}_j \vec{j} + \{ab\}_k \vec{k}$.
Finally:
3:the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ have equal magnitudes:$| \vec{i} | = | \vec{j} | = | \vec{k} |$.
As a consequence, the "mathematical trick" of expressing the cross product vector $\vec{a} \times \vec{b}$ as the above determinant "works":
The component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{a}$ vanishes explicitly:
$$\left( a_i (a_j b_k - a_k b_j) \frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_j (a_k b_i - a_i b_k) \frac{(| \vec{j} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_k (a_i b_j - a_j b_i) \frac{(| \vec{k} |)^2}{| \vec{a} \times \vec{b} |} \right) = $$
$$\frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \left( a_i (a_j b_k - a_k b_j) + a_j (a_k b_i - a_i b_k) + a_k (a_i b_j - a_j b_i) \right) = 0,$$
and likewise the component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{b}$ vanishes explicitly;
i.e. cross product vector $\vec{a} \times \vec{b}$ is expressed explicitly orthogonal to both vectors $\vec{a}$ and $\vec{b}$.
And, no less important the magnitude of cross product vector $\vec{a} \times \vec{b}$ "comes out correctly", i.e. such that
$$ \Big( | \vec{a} \times \vec{b} | \Big)^2 := $$ $$ \left( (a_j b_k - a_k b_j)^2 + (a_k b_i - a_i b_k)^2 + (a_i b_j - a_j b_i)^2 \right) ~ (| \vec{i} |)^4 = $$ $$ \left( (a_i^2 + a_j^2 + a_k^2) ~ (b_i^2 + b_j^2 + b_k^2) \right) ~ (| \vec{i} |)^4 - \left( (a_i b_i + a_j b_j + a_k b_k) ~ (| \vec{i} |)^2 \right)^2 := $$ $$ \Big( | \vec{a} | \Big)^2 \Big( | \vec{b} | \Big)^2 - | \vec{a} | ~ | \vec{b} | ~ a_b ~ b_a,$$
where $b_a$ denotes the component of vector $\vec{b}$ "along/parallel to" vector $\vec{a}$, and $a_b$ denotes the component of vector $\vec{a}$ "along/parallel to" vector $\vec{b}$.