Intuition#
If Dot product, is interactions between similar dimensions, cross product is interactions between different dimensions.
For two vectors \(\vec{a}\) and \(\vec{b}\), we can write every combination of their components in a grid:
$$ \vec{a} = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix} $$
$$ \begin{array}{c|ccc} & b_x & b_y & b_z \\ \hline a_x & \text{Dot} & \text{Cross} & \text{Cross} \\ a_y & \text{Cross} & \text{Dot} & \text{Cross} \\ a_z & \text{Cross} & \text{Cross} & \text{Dot} \end{array} $$
You can also think of cross product as difference product, what’s not in common product.
Why do we need to think about differences?
- Think about two vectors parallel in a plane, we do not have area between them as they are just two parallel lines.
- But if we have a perpendicular vector to a vector now there is some area between them.
- So, area is only present when vectors are going in different dimensions.
Calculations#
Cross product of two vectors \(\vec{v}\) and \(\vec{w}\) is the area of the parallelogram it forms:

Almost, we also have to take care of the direction:
- If, \(\vec{v}\) is rotated towards \(\vec{w}\) in a counter-clockwise direction,
the cross product is positive.

- If, \(\vec{v}\) is rotated towards \(\vec{w}\) in a clockwise direction,
the cross product is negative.

NOTE: Think of the right-hand rule to determine the direction of the cross product.
- Notice, this means the order of the vectors matters: So, \(\vec{v} \times \vec{w} = - (\vec{w} \times \vec{v})\)
- If, \(\vec{v}\) is rotated towards \(\vec{w}\) in a counter-clockwise direction,
the cross product is positive.
Determinant is used to calculate the cross product, because we can think of the area to be what happens when we apply the linear transformation to the \( \hat{i} \) and \( \hat{j} \) unit vectors.

Example: Consider, two vectors: $$ \vec{v} = \begin{bmatrix} -3 \\ 1 \end{bmatrix}, \quad \vec{w} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $$ $$ \vec{v} \times \vec{w} = \text{det}\begin{bmatrix} -3 & 2 \\ 1 & 1 \end{bmatrix} = (-3)(1) - (2)(1) = -3 - 2 = -5 $$
One thing to note: If two vectors are more perpendicular to each other, the area between them is larger, and hence the cross product is larger. And if they are more parallel to each other, the area between them is smaller, and hence the cross product is smaller.

Another thing: If you scale one of the vectors, the area between them also scales by the same factor.
So, here:
$$
(3 \cdot \vec{v}) \times \vec{w} = 3 \cdot (\vec{v} \times \vec{w})
$$The cross product will give you another vector that is perpendicular to both \(\vec{v}\) and \(\vec{w}\) in 3D space, the area is just the magnitude of that vector.

Right-hand rule

Formula:
$$ \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix} \times \begin{bmatrix} w_x \\ w_y \\ w_z \end{bmatrix} det \begin{bmatrix} \hat{i} & v_x & w_x \\ \hat{j} & v_y & w_y \\ \hat{k} & v_z & w_z \end{bmatrix} $$
$$ = \hat{i}(v_y w_z - v_z w_y) - \hat{j}(v_x w_z - v_z w_x) + \hat{k}(v_x w_y - v_y w_x) $$