# Quick Math Cross Product Matrix

## Quick Math Cross Product Matrix

1
##### Enter expression in this form

Example (3x3 Matrix):  Enter expression as [3,4,5],[6,7,8]

### Cross Product Matrix

In vector algebra, the cross product is an operation on two vectors in three-dimensional space, producing another vector that is perpendicular to the plane containing the original vectors. This can be represented using a special matrix called the cross product matrix or skew-symmetric matrix.

#### Definition

Given a vector $$\mathbf{a} = [a_1, a_2, a_3]^T$$, the cross product matrix $$[ \mathbf{a} ]_\times$$ is defined as:

$[\mathbf{a}]_\times = \begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{pmatrix}$

#### Cross Product Using the Matrix

For two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$, the cross product $$\mathbf{a} \times \mathbf{b}$$ can be computed using the cross product matrix as follows:

$\mathbf{a} \times \mathbf{b} = [\mathbf{a}]_\times \mathbf{b}$

#### Example

Let’s compute the cross product of $$\mathbf{a} = [1, 2, 3]^T$$ and $$\mathbf{b} = [4, 5, 6]^T$$.

1. Construct the Cross Product Matrix:

$[\mathbf{a}]_\times = \begin{pmatrix} 0 & -3 & 2 \\ 3 & 0 & -1 \\ -2 & 1 & 0 \end{pmatrix}$

2. Compute the Cross Product:

$\mathbf{a} \times \mathbf{b} = [\mathbf{a}]_\times \mathbf{b} = \begin{pmatrix} 0 & -3 & 2 \\ 3 & 0 & -1 \\ -2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$

Perform the matrix-vector multiplication:

$= \begin{pmatrix} 0 \cdot 4 + (-3) \cdot 5 + 2 \cdot 6 \\ 3 \cdot 4 + 0 \cdot 5 + (-1) \cdot 6 \\ -2 \cdot 4 + 1 \cdot 5 + 0 \cdot 6 \end{pmatrix} = \begin{pmatrix} 0 – 15 + 12 \\ 12 + 0 – 6 \\ -8 + 5 + 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$

#### Verification

To verify, we can use the standard cross product formula:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = \mathbf{i}(2 \cdot 6 – 3 \cdot 5) – \mathbf{j}(1 \cdot 6 – 3 \cdot 4) + \mathbf{k}(1 \cdot 5 – 2 \cdot 4) = \mathbf{i}(-3) – \mathbf{j}(-6) + \mathbf{k}(-3) = [-3, 6, -3]$

Thus, the cross product calculated using the cross product matrix is verified. This example illustrates how to compute the cross product of two vectors using the cross product matrix approach.