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

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?

• Agha, your question has already attracted one Move to MathSE vote, so I've tried to rewrite it in a way more applicable to the PhysicsSE. However this is a big change. If you don't like my change please roll it back. – John Rennie Aug 31 '14 at 5:44
• There's a very nice post on math.SE: What's an intuitive way to think about the determinant? Take a look at how the properties of the determinant uniquely determine its mathematical form. I suspect this argument can be related to the cross product, but it's late here. – BMS Aug 31 '14 at 9:56
• I always look at the cross product as a linear operator $$\vec{a}\times\vec{b} = \begin{pmatrix}0&-a_k&a_j\\a_k&0&-a_i\\-a_j&a_i&0\end{pmatrix}\begin{pmatrix}b_i\\b_j\\b_k\end{pmatrix}$$ – John Alexiou Aug 31 '14 at 18:00
• @ja72 i am not from a mathematical background, and i have no idea about a LINEAR OPERATOR, will you please elaborate – agha rehan abbas Aug 31 '14 at 18:02
• What I mean is that the $[\vec{a}\times]$ matrix transforms the $\vec{b}$ vector. For example in a rotating frame $[\vec{\omega}\times] \vec{r}$ returns the rate of change of the components of a body fixed vector $\vec{r}$. The $\times$ takes a vector and turns it into an operator (like a function but in vector space). – John Alexiou Aug 31 '14 at 22:32

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 last term in the last formula, $(a_bb_a)^2$, seems to be a mistake. By your definition $a_b=a\cos\alpha$ and $b_a=b\cos\alpha$ with $\alpha$ being the angle between vectors, so $a_bb_a=ab\cos^2\alpha$, the power of cosine being 1 more than needed. – firtree Aug 31 '14 at 14:30
• firtree: "The last term in the last formula, [...] seems to be a mistake." -- That's right, thanks! Explicitly: $$\Big(|\vec{a}\times\vec{b}|\Big)^2:=$$ $$\left((a_j b_k-a_k b_j)^2+(a_k b_i-a_i b_k)^2+(a_i b_j-a_j b_i)^2\right)~(|\vec{i}|)^4=$$ $$\left((a_i^2+a_j^2+a_k^2)~(b_i^2+b_j^2+b_k^2)\right)~(|\vec{i}|)^4-\left((a_i b_i+a_j b_j+a_k b_k)~(|\vec{i}|)^2\right)^2:=$$ $$\Big(|\vec{a}|\Big)^2~\Big(|\vec{b}|\Big)^2-|\vec{a}|~| \vec{b}|~a_b~b_a;$$ where I've tried hard, of course, not to introduce any notions (such as "angle") which don't appear in the question itself. – user12262 Aug 31 '14 at 21:10
• What do the braces terms (eg $\lbrace ab\rbrace_i$) stand for? – Stan Shunpike Feb 18 '15 at 6:58
• @Stan Shunpike: "What do the braces terms (eg $\{ab\}_i$) stand for?" -- This was just some ad hoc ("seat-of-my-pants") notation for expressing one particular "component coefficient" (real or complex number) of the cross product $\vec a \times \vec b$, referring to the (chosen) basis vector $\vec i$. It was meant to formally express that this "component coefficient" depends on vectors $\vec a$, and $\vec b$, and $\vec i$; without already writing down an explicit expression. Actually, for orthogonal basis vectors: $$\{ab\}_i := (a_j~b_k - a_k~b_j)~\frac{|\vec j|~|\vec k|}{|\vec i|},$$ etc. – user12262 Feb 18 '15 at 22:16

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)

• I think it's important to say "determinant" (with quotes) since in general a determinant is a scalar value – information_interchange Jun 13 '18 at 4:35

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.

Credit to @jimmyz for the first mention of the area interpretation.

Since any two vectors can be put in a plane, we can look at the two-dimensional case without losing too much of our detail. The cross product of these two vectors gives the area of this parallelogram.

Usually, I think of the one coming out of the origin, and then the one that comes from that vector, but any two vectors in order will give the same result.

In this case, $$A=\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}=2 \cdot 3-1 \cdot 1=5$$

We can break this down into simple cases with the diagonal and off-diagonal components. $$A=\begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix}=2 \cdot 3-0 \cdot 0=6$$

And if we look at the other components of the vector, we get a negative area. $$A=\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix}=0 \cdot 0-1 \cdot 1=-1$$

What helped me to understand how determinant is linked to cross product is to consider components of a determinant:

$$\begin{vmatrix} a_x & b_x & c_x\\ a_y & b_y & c_y\\ a_z & b_z & c_z\\ \end{vmatrix} = c_x \begin{vmatrix} a_y & b_y\\ a_z & b_z\\ \end{vmatrix} - c_y \begin{vmatrix} a_x & b_x\\ a_z & b_z\\ \end{vmatrix} + c_z \begin{vmatrix} a_x & b_x\\ a_y & b_y\\ \end{vmatrix}$$

The cool thing is that the 2x2 determinants are the projection areas of the parallelogram formed by $$\vec a$$ and $$\vec b$$ onto $$yz$$, $$xz$$ and $$xy$$ plains. So if you sum the squares of the projection areas and take square root you'll get the area of the parallelogram formed by $$\vec a$$ and $$\vec b$$ (The Full Pythagorean Theorem):

$$\sqrt{ \begin{vmatrix} a_y & b_y\\ a_z & b_z\\ \end{vmatrix}^2 + \begin{vmatrix} a_x & b_x\\ a_z & b_z\\ \end{vmatrix}^2 + \begin{vmatrix} a_x & b_x\\ a_y & b_y\\ \end{vmatrix}^2 }$$

i.e. the length of a vector with these components (2x2 determinants) is equal to the area of the parallelogram formed by $$\vec a$$ and $$\vec b$$.

$$\vec v = \begin{pmatrix} \begin{vmatrix} a_y & b_y\\ a_z & b_z\\ \end{vmatrix}\\ -\begin{vmatrix} a_x & b_x\\ a_z & b_z\\ \end{vmatrix}\\ \begin{vmatrix} a_x & b_x\\ a_y & b_y\\ \end{vmatrix} \end{pmatrix}$$

What is left to show is that this vector is perpendicular to both $$\vec a$$ and $$\vec b$$. If you calculate the dot products $$\vec v \cdot \vec a$$ and $$\vec v \cdot \vec b$$ you'll verify that they both are equal to zero.

To make it easy to see the logic behind, imagine that the first vector is directed to $\vec{i}$ and the second to $\vec{j}$. You get

$$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & 0 & 0 \\ 0 & b_j & 0 \end{matrix}\right|$$

and according to Wikipedia the product in this case is $ab\vec{k}$.

There are some physical phenomena which can be described with vector products such as electromagnetic induction. More generally is the example where two forces in space act on a body and no one is directed to the axis of your coordinate system. Of course you are free to rotate your coordinate system to one of the forces and you get an improved calculation, that has more zeros in the determinant.

I'm going to give my answer here since I'm not particularly satisfied with the others:

The "cross product as determinant" is just a trick; there is no deeper meaning or relationship between the two. It just so happens that when I define a special, weird matrix with basis vector and target vectors $a$ and $b$ all jumbled together, then the "determinant" of that matrix is an expression which happens to be the cross product. In particular note that the determinant generally is a scalar value, and here we bend the rules to reinterpret the determinant as a vector value.

See ("formal calculation" here) and this good Quora answer

The most succinct proof uses the Levi-Civita symbol: the vector product can be defined as $\epsilon_{ijk}\vec{e}_i a_j b_k$, which by inspection is the desired determinant. That may make it look like a coincidence, but we can get more insight than that. The reason the LCS appears in both contexts is because it's fully antisymmetric.

Given $n-1$ vectors $a_1,\,\cdots a_{n-1}$ in $K^n$ for some field $K$, we seek an $n$th vector orthogonal to each. (That's your physical motivation, if you need one.) That vector is $\epsilon_{i_n i_1\cdots i_{n-1}}\vec{e}_{i_n}(a_1)_{i_1}\cdots (a_{n-1})_{i_{n-1}}$, because for $1\le j\le n-1$ the dot product with $\sum_{k} (a_j)_k \vec{e}_k$ just replaces $\vec{e}_{i_n}$ with $(\vec{a_j})_{i_n}$, giving the determinant of a vector with two equal rows.