Formula for cross product

Question

The formula for the cross product of two vectors in $R^3$, $\vec{a} = (a_1, a_2, a_3)$ and $\vec{b} = (b_1, b_2, b_3)$ is $$\det\begin{pmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{pmatrix}$$

I know that in general for three 3D vectors the determinant represents the volume of the parallelepiped. But how is it valid to put (basis) vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ into a vector, and what graphical/intuitive significance does it have? What would have been the initial motivation of this formula?

Note: I already read through similar questions and corresponding answers but was not satisfied. So please do not downvote this question and if possible give me some insight.

Answer 1

I have linked a video in the comments above, but I'll summarize the ideas from the video in this answer. We have 2 questions to address:

1. Why does it make sense to put $\mathbf{i,j,k}$ into a determinant?
2. How does this result in a vector with the geometric properties of the cross-product?

To begin, I claim that the "natural" form of the cross-product is not that of a vector, but instead that of a function. Given two vectors $\mathbf a, \mathbf b$, the cross-product is really the function $f_{\mathbf a,\mathbf b}: \Bbb R^3 \to \Bbb R$ defined by $$ f_{\mathbf a,\mathbf b}(\mathbf x) = \det\pmatrix{x_1 & x_2 & x_3\\a_1 & a_2 & a_3\\ b_1 & b_2 & b_3}. $$ In other words, $f_{\mathbf a, \mathbf b}$ is a function that takes as its input a vector $\mathbf x = (x_1,x_2,x_3)$, and as its output produces the (signed) volume of the parallelpiped with edges $\mathbf{a,b,x}$. It is easy to think of this function as a vector because it is a linear function, and every linear function $f:\Bbb R^3 \to \Bbb R$ can be written in the form $$ f(\mathbf x) = p_1 x_1 + p_2 x_2 + p_3 x_3. $$ (Note that we must use “signed” volume in order for this to be the case. If $f$ only produced positive numbers, it would be impossible to have $f(-\mathbf x)=-f(\mathbf x)$ as the above form implies).

The vector $\mathbf p = (p_1,p_2,p_3)$ can be thought of as a way of "encoding" the function $f$, since we have $f(\mathbf x) = \mathbf p \cdot \mathbf x$, where $\cdot$ denotes a dot-product. This leads to our answer to question 1: when we compute the determinant with $\mathbf {i,j,k}$ has the top-row entries, we end up with a vector $p_1 \mathbf i + p_2 \mathbf j + p_3 \mathbf k$. The vectors $\mathbf{i,j,k}$ act as place-holders for our input-coordinates $x_1,x_2,x_3$: if we calculate $\mathbf p \cdot \mathbf x$, we end up with $$ p_1 x_1 + p_2 x_2 + p_3 x_3, $$ which is exactly what we would have gotten if we calculated the determinant with the entries of $\mathbf x$ as our top row, i.e. if we calculated $f_{\mathbf a, \mathbf b}(\mathbf x)$ from the definition I give above.

As for question 2, we can make the following observations:

• If $\mathbf x$ is either $\mathbf a$ or $\mathbf b$, then $\mathbf p \cdot \mathbf x = f_{\mathbf a, \mathbf b}(\mathbf x) = 0$. So, $\mathbf p$ is orthogonal to both $\mathbf a$ and $\mathbf b$.
• If we plug in $\mathbf x = \mathbf p/\|\mathbf p\|$, then we should find that $\mathbf p \cdot \mathbf x = \frac{\mathbf p \cdot \mathbf p}{\|\mathbf p\|} = \|\mathbf p\|$. From the positivivity of the determinant, we see that the direction of $\mathbf p$ is such that $\mathbf p, \mathbf a, \mathbf b$ forms a right-handed system. From the fact that $f_{\mathbf a, \mathbf b}(\mathbf x) = \|\mathbf p\|$, we can deduce that $\|\mathbf p\|$ is equal to the area of the parallelogram with edges $\mathbf a, \mathbf b$.

With this, we deduce the properties of the cross-product: $\mathbf p = \mathbf a \times \mathbf b$ is always orthogonal to $\mathbf a , \mathbf b$. When $\mathbf p \neq \mathbf 0$, $\mathbf p$ points in the direction such that $\mathbf a, \mathbf b, \mathbf p$ forms a right-handed basis. Finally, $\|\mathbf p\|$ is the area of the parallelogram with sides $\mathbf a, \mathbf b$.

Answer 2

As several comments have pointed out, what you have written is a mnemonic device, not an actual formula. Here is a real formula, in the same spirit. $$\eqalign{ \vec a\times \vec c &= &{\bf i}\,&\det\Big(\,\big[\matrix{\bf i&\vec a&\vec c}\big]\,\Big) \\ &+ \;&{\bf j}\,&\det\Big(\,\big[\matrix{\bf j&\vec a&\vec c}\big]\,\Big) \\ &+ &{\bf k}\,&\det\Big(\,\big[\matrix{\bf k&\vec a&\vec c}\big]\,\Big) \\ }$$

Answer 3

For anyone that wants explanation with graphs and figures, I post the way how i understand cross product in my website. Hope it helps!

https://www.schoolingerscat.com/math/crossProduct/introduction

The summary of my approach is:

1. We first need to understand that determinant is the oriented volume of a given full-rank matrix.

2. Next we show that if we fix $\vec{a}$ and $\vec{b}$ and make $\vec{v}$ as a variable vector. we then can create a linear functional $\phi$ such that:

$$ \phi(\vec{v}) = \det \begin{bmatrix} v_1 & v_2 & v_3 \\ a_1 & a_2 & b_2 \\ b_1 & a_3 & b_3 \end{bmatrix} $$

3. Then we can show that there is a unique vector, $\vec{p}$, such that $\phi(\vec{v})= \vec{p} \cdot \vec{v} $

4. Next we prove that $\vec{p} = \vec{a} \times \vec{b}$ by showing that $\vec{p}$ is aligned with unit vector $\hat{u}$ that is orthogonal to Area formed by $\vec{a},\vec{b} $ and also show $\| \vec{p} \| = \text{ Area formed by } \vec{a} \text{ , } \vec{b}$

5. Finally, with $\vec{p} = \vec{a} \times \vec{b}$, and $ \phi(\vec{v}) = \vec{p} \cdot \vec{v} =\det \begin{bmatrix} v_1 & v_2 & v_3 \\ a_1 & a_2 & b_2 \\ b_1 & a_3 & b_3 \end{bmatrix} $, we can find $ p_1 , p_2 , p_3$ of $\vec{p}$ with $ p_1 = \vec{p} \cdot \hat{i} , p_2 = \vec{p} \cdot \hat{j} , p_3 = \vec{p} \cdot \hat{k}$ , which if you put them together, will give you the "notation": $$\vec{p} = \det \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & b_2 \\ b_1 & a_3 & b_3 \end{bmatrix} = \vec{a} \times \vec{b} $$

Answer 4

You can use the linear algebra form of the cross-product

$$ \boldsymbol{a} \times \boldsymbol{b} = [\boldsymbol{a}\times] \boldsymbol{b} $$

where

$$ [\boldsymbol{a}\times] = \begin{bmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \end{bmatrix} $$

and

$$\boldsymbol{b} = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}$$

Expanded out this is

$$ \begin{bmatrix} a_x\\ a_y \\ a_x \end{bmatrix} \times \begin{bmatrix} b_x\\ b_y \\ b_x \end{bmatrix} = \begin{bmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \end{bmatrix} \begin{bmatrix} b_x\\ b_y \\ b_x \end{bmatrix} = \begin{bmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end{bmatrix} $$

This skew-symmetric matrix form of the cross-product operator is very useful.

Since it is just linear algebra it also obeys the associative property. For example

$$ \boldsymbol{a} \times ( \boldsymbol{b} \times \boldsymbol{c} ) = [\boldsymbol{a}\times] [\boldsymbol{b}\times] \boldsymbol{c} = [\boldsymbol{a}\times] ([\boldsymbol{b}\times] \boldsymbol{c}) = ( [\boldsymbol{a}\times] [\boldsymbol{b}\times]) \boldsymbol{c} $$

and also has the following triple product property

$$ ( \boldsymbol{a} \times \boldsymbol{b}) \times \boldsymbol{c} = \left( [\boldsymbol{a}\times] [\boldsymbol{b}\times] - [\boldsymbol{b}\times] [\boldsymbol{a} \times] \right) \boldsymbol{c} $$