Why can the cross product of two vectors be calculated as the determinant of a matrix?

Is there any physical significance to this matrix

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 is

• either 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 $\vec{b}$.

The cross product is defined to be the vector which is perpendicular to both vectors, so for instance the force exerted on a rod moving in a magnetic field is perpendicular to both its velocity and the field, hence is given by their cross product.

Now if you work out which vector is perpendicular to both vectors you get the determinant of the two vectors (In other words, writing it as a determinant is only to make it more easy on the eye, I don't think there is a deep reason to it)

No, this is NOT a trick. Your intuition was quite correct in thinking there might be some geometric significance to this. I’ll give you a broad hint, and then you can have the pleasure of figuring the details out yourself. Here is the hint that should give you the geometric insight into why this works out this way:

The Cross Product is a vector operation that produces a vector whose magnitude is defined by an area formed by the two input vectors. The determinate of a matrix is also related to an area formed from the vectors.

I am a little surprised that none of the obviously talented math boffins who replied earlier pointed out this obvious geometric interpretation. I believe this was closer to the mathematical intuition that you were looking for in your original question.

P.S. I will try to fill out this answer, with the actual argument, including a numerical example, if I get a chance.