[Linear Algebra] 9. Cross products via transformations

2 dimension

The cross product of $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$, written $\mathbf{\vec{v}} \times \mathbf{\vec{w}}$, is the area of the parallelogram defined by $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$. Cross product also considers orientation. 1). If $\mathbf{\vec{v}}$ is on the right of $\mathbf{\vec{w}}$, then $\mathbf{\vec{v}} \times \mathbf{\vec{w}}$ is positive and equal to the area of parallelogram. 2). But if $\mathbf{\vec{v}}$ is on the left of $\mathbf{\vec{w}}$, then the cross product is negative, namely the negative area of the parallelogram.

 

To compute the 2-D cross product $\mathbf{\vec{v}} \times \mathbf{\vec{w}}$, 1). make $2 \times 2$ matrix which first column is the coordinates of $\mathbf{\vec{v}}$ and second column is the coordinates of $\mathbf{\vec{w}}$. 2). compute the determinant.

$$
\mathbf{\vec{v}} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}
\ 
\mathbf{\vec{w}} = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix}
\qquad
\mathbf{\vec{v}} \times \mathbf{\vec{w}} =
\det
\begin{pmatrix}
\begin{bmatrix}
v_1 & w_1 \\
v_2 & w_2 \\
\end{bmatrix}
\end{pmatrix}
= v_1w_2 - w_1v_2
$$

 

The determinant is a factor representing "how much the area of single unit square changes".

So 2-D cross product is a determinant which is the area of the parallelogram.

3 dimension

Cross product described by 2-D vectors is not technically correct. The true cross product is something that combines two different 3-D vectors to get a new 3-D vector. Just as before, parallelogram defined by two vectors play a big role. The new vector's length will be the area of the parallelogram. And the direction of new vector is going to be perpendicular to the parallelogram. Orientation of new vector is decided by right-hand rule. 1). Put the forefinger of your right hand in the direction of $\mathbf{\vec{v}}$ then 2). stick out your middle finger in the direction of $\mathbf{\vec{w}}$. 3). Then when you point up your thumb, that's the direction of the cross product.

 

3-D cross product formula looks truly strange at first. 1). Defines $3 \times 3$ matrix where the second and third columns are $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$, but for the first column is basis vectors $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$, and $\mathbf{\hat{k}}$. 2). Then compute the determinant of this matrix. And the vector defined by that linear combination is the unique vector perpendicular to $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$, whose size is the area of the parallelogram and whose direction obeys the right-hand rule.

$$
\mathbf{\vec{v}} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}
\ 
\mathbf{\vec{w}} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}
$$

$$
\mathbf{\vec{v}} \times \mathbf{\vec{w}} =
\det
\begin{pmatrix}
\begin{bmatrix}
\mathbf{\hat{i}} & v_1 & w_1 \\
\mathbf{\hat{j}} & v_2 & w_2 \\
\mathbf{\hat{k}} & v_3 & w_3 \\
\end{bmatrix}
\end{pmatrix}
= \mathbf{\hat{i}}(v_2w_3 - v_3w_2) + \mathbf{\hat{j}}(v_3w_1 - v_1w_3) + \mathbf{\hat{k}}(v_1w_2 - v_2w_1)
$$

 

In some sense, this is just a notational trick. But there is a reason for doing it. To understand where all of this comes from it helps to use the idea of duality.