Intuition behind cross-product and area of parallelogram

Question

The cross product in 2D is defined like that: $|(x_1, y_1) \times (x_2, y_2)| = x_1 y_2 - x_2 y_1.$

I perfectly understand the first part of the definition: $x_1 y_2$, which is simply the area of a rectangle:

I am struggling to understand the second part: $- x_2 y_1.$
I feel that the second part, probably, has to do with the rotations of the vectors $(x_1, y_1)$ and $(x_2, y_2)$, because when we rotate the original rectangle we should preserve the area and $- x_2 y_1$ somehow compensates for the excess amount of area that we get from the first term $x_1 y_2$.

I feel that my intuition lacks a lot of details and I would be grateful for the explanations of the second term and it's connection to the rotations.

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My question is different from Reasoning behind the cross products used to find area, although the titles are almost identical. The orthogonality of my question to that can be seen, by reading these parts of the question:

I do not have any issues with calculating the area between two vectors.

But I have.

...but not why the cross product is used instead of the dot product.

The dot product is out of the scope of my question.

Answer 1

• The formula works fine for the standard unit vectors.
• Stretching one of the vectors by a constant $c$ should multiply the (oriented) area by $c$ and does indeed multiply the cross product by $c$
• Shearing along $(x_1,y_1)$, i.e., replacing $(x_2,y_2)$ with $(x_2+cx_1,y_2+cy_1)$ does not change the area and also does not change the cross product (the extra terms cancel)

As any parallelogram can be obtained from the standard unit vectors by a few steps of shearing/stretching, the cross product tells us the oriented area for all parallelograms.

Answer 2

The cross product can also be defined as $a\times b=|a||b|\sin\theta$, with $\theta$ as the angle between $a$ and $b$.

It can be seen as giving a vector perpendicular to both $a$ and $b$, with length $|a||b|\sin\theta$, or as the area of a skew-rectangle (i.e. a parallelgram) with side lengths $a$ and $b$ and a skew of $\theta$.

Answer 3

If you prefer a geometrical vision, lock at the figure.

The area of the parallelogram $OMPN$ is obviously the same as the area of the pink parallelogram (same basis $OM$ and same height because the opposite side are parallel). But the area of the pink parallelogram is also given by:$ \quad Area=|a_1c|$ and we can find $c$ as the intersect at the origin of the line from $N,P$.

Whith a bit of elemental analitic geometry you can find that $$ c=\frac{a_1b_2-a_2b_1}{a_1} $$

and this menas that:

$$ Area= |a_1b_2-a_2b_1| $$

that is the absolute value of the derminat of the matrix that has as columns the two vectors $\overline {OM}$ and $\overline {ON}$ or the absolut value of the cross product of the same two vectors.

Answer 4

Here is a presentation of the determinant on all-positive quantities which can provide a first-level understanding of the connection with areas.