Rather than using the dot product equation, let's use the electric flux equation  without the dot product.

我们点积方程，而是带点积电通量方程。

Electric flux equals the integral of the dot product of electric field and dA.

电通量等于电场与dA点积积分。

However, let's use E dA cosine theta instead of the dot product.

但是，让我们 E dA 余弦西塔来代替点积。

Electric flux equals the dot product of the electric field and the area, so let's  use that.

电通量等于电场与面积点积，所我们就它。

Solving a linear system with an orthonormal matrix is actually super easy, because dot products are preserved.

标准正交矩阵求解线性方程组其实非常简单 因为点积被保留了。

That is, they don't preserve that zero dot product.

也就是说 它们保留0点积。

The dot product before and after the transformation will look very different.

变换前后点积看起来很一样。

And looking at the example I just showed, dot products certainly aren't preserved.

看看我刚才举例子 点积肯定会被保留。

The relevant background here is understanding determinance, little bit of dot products, and of course, linear systems of equations.

背景知识是理解行列式 一点点积 当然还有线性方程组。

In the sense that applying the transformation is the same thing as taking a dot product with that vector.

在这个意义上应这个变换就当于对这个向量做点积。

In fact, worthwhile side note here transformations which do preserve dot products are special enough to have their own name.

事实上 值得注意是保留点积变换很特别 有自己名字。

And if they point in generally the opposite direction, their dot product is negative.

如果它们通常指向反方向 它们点积是负。

The excordinate of this mystery input vector is what you get by taking its dot product with the 1st basis vector one zero.

这个神秘输入向量纵坐标就是它和第一个基向量(0)点积。

Likewise, things that start off perpendicular with dot product zero, like the two basis factors, quite often don't stay perpendicular to each other after the transformation.

同样地 开始与点积0垂直东西 比如两个基因子 在变换之后通常会互垂直。

When their perpendicular meaning, the projection of one onto the other, is the zero vector, their dot product is zero.

当它们垂直意思 一个向量在另一个向量上投影 是零向量时 它们点积是零。

And that's probably the most important thing for you to remember about the dot product.

这可能是你们要记住于点积最重要一点。

Looking at sides 2 and 4, we need to realize the dot product of B and ds, is the same as B ds cosine theta, Right!

观察边2和边4， 我们需要实现B和ds点积，与B ds cosine theta同， 对吧！

So that wraps up dot products and cross products.

这就包含了点积和叉乘。

So that performing the linear transformation is the same as taking a dot product with that vector the cross product.

所进行线性变换就等于对这个向量做点积也就是叉乘。

Now, this numerical operation of multiplying a one by two matrix by a vector feels just like taking the dot product of two vectors.

现在 这个1×2矩阵乘一个向量数值运算就像取两个向量点积。