# Expressing area of a parallelogram as sum of areas of two rectangles using determinant

Say A= $\begin{bmatrix} x_{1} \\ y_{1} \\ \end{bmatrix}$ and $B=\begin{bmatrix} x_{2} \\ y_{2} \\ \end{bmatrix}$ are two perpendicular vectors with unit length.
I've been trying to understand how the expression $\color{red}{x_1y_2}-\color{grey}{x_2y_1}$ gives the area of the parallelogram formed by these vectors. I've started with perpendicular unit vectors for simplicity.

Ofcourse area of the blue square doesn't change under rotation, but I'm wondering if it's possible to see why the continuously changing areas of red and black squares add up to the area of blue square ?

By looking at the $x$-axis, in your diagram it can be clearly seen that the width of the red and black squares are defined by the $x$- coordinate of the $\pmb{A}$ vector and the $\pmb{B}$ vector, respectively.

Thus, the side length of the red square is $\cos(a)$.

Similarly, the side length of the black square is $\cos(180-a-90) = \cos(90-a) = \sin(a)$.

Thus, the area of the red square is $\cos^2 a$ and the area of the black square is $\sin^2 a$.

Using the identity $\cos^2a + \sin^2a =1$,

it follows that the sum of the red square and the black square is unity.

• Brilliant! As the angle $a$ increases, $\cos a$ decreases and $\sin a$ increases keeping the sum $\cos^2a + \sin^2a$ constant. I'll see if I can figure something like this for perpendicular non unit vectors. Thank you :) – rsadhvika Jul 12 '18 at 9:53

$x_2=-y_1$ and $y_2=x_1$ so that $x_1y_2-x_2y_1=x_1^2+y_1^2=x_2^2+y_2^2=x_1^2+x_2^2=y_1^2+y_2^2$, the constant area.

The animation also illustrates the theorm of Pythagoras,

$$x_1^2+y_1^2=x_1^2+x_2^2=a^2,\\x_2^2+y_2^2=x_2^1+x_1^2=b^2.$$

• Wow! By same reasoning, when the vectors are not unit, $(x_2,y_2) = (-ky_1,kx_1)$; so $x_1y_2-x_2y_1 = k(x_1^2+y_1^2)$, the constant area. You're awesome! Thank you so much :) I'll try the general case now where the vectors are not perpendicular. – rsadhvika Jul 12 '18 at 10:06
• Ahh sneaky pythagoras! so here the projections of $a$ or $b$ onto $x, y$ axes are the legs and the vectors $a,b$ themselves are the hypotenuse. Nice nice ! – rsadhvika Jul 12 '18 at 10:11