Why is the parallelogram area between two vectors (x1, Y1), (X2, Y2) (x1 * y2-x2 * Y1 )?

Link: http://blog.csdn.net/zhangxaochen/article/details/8027003

All of a sudden, I want to solve this problem. In fact, this is just a sentence that I don't know ..

It is estimated that college students like me have to wait for a moment to see this question...

At least I forgot about the determining factor.

 

1. I first thought of High School analytic ry knowledge. Because if you think of (X2, Y2) as the bottom edge of Parallelogram, the side length is SQRT (X2 * X2 + y2 * Y2 ). Then, calculate the distance from the vertex (x1, Y1) to the straight line {(), (X2, Y2, D = | x1y2-x2y1 |/SQRT (X2 * X2 + y2 * Y2 ). Then the base multiplied by height is the x1y2-x2y1.

 

2. But the above solution is neither clever nor concise. IRC # math channel is well-intentioned and provides a junior high school answer version, which is clear at a glance:

The blue parallelogram is the area to be calculated. The area of the Area 1, 2, and 3 can be reduced by a pink right triangle, and the area of the entire parallelogram is quickly obtained.

 

3. The subject should have been the deciding factor and linear ing. I just forgot it and didn't pick it up again. So no.

Leave a comment on CALE:

If you know already that if the area of some set S is a, and T is a linear map, then t (S) = {T (V ): V in S} has area | det T | A, then you just note that the parallelogram spanned by (x1, Y1) and (X2, Y2) is the image of the Unit
Square, I. e. the parallelogram spanned by (1, 0) and (0, 1) under the linear map whose matrix representation consists of those two vectors

 

Think about the effect. For example, in Square, in the upper right corner (), the vector (x1, Y1) (X2, Y2) is mapped to the coordinate (X1 + X2, y1 + y2)

Only theory can be supplemented first ~~

Link: http://blog.csdn.net/zhangxaochen/article/details/8027003

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