Because the Gaussian surface we chose was a sphere concentric with the point charge, the  electric field and the area vector were always in the same direction giving an angle of zero degrees  in the dot product.

因为所选的高斯面是一个与点电荷同心的球体，电场和面积矢量总是同方的， 在点乘中角度为零度。

Since we just saw that this change vector has a constant magnitude, the only way for this dot product to stay the same is if the angle it makes with that collision line stays the same.

由于到这个变化量的大小是恒定的，点积保持不变的唯一方法就是它与碰撞线形成的夹角保持不变。

E.g. you could have two vectors, generally pointing in the same direction with a positive dot product, which get pulled apart from each other during the transformation, in such a way that they end up having a negative dot product.

例如 你可以有两个量 它通常指同一个方 点积为正 在变换过程中它被拉开 以这样的方式 它最终得到负的点积。

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

然，使用E dA 余弦 theta 不是点乘。

For example, you could have two vectors generally pointing in the same direction, with a positive dot product, which get pulled away from each other during the transformation, in such a way that they then have a negative dot product.

例如， 你可以有两个大致指同一方的量，它的点积为正， 在变换过程中它被拉开，以至于它的点积变为负。

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

电通量等于电场与dA的点积的积分。

So when two vectors are generally pointing in the same direction, their dot product is positive.

所以当两个量通常指同一个方时 它的点积是正的。

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

事实上 值得注意的是保留点积的变换很特别 有自己的名字。

Notice, this looks like a dot product between two column vectors, [m1, m2], and [v1, v2].

注意，这起来像是两个列量 [m1, m2] 和 [v1, v2] 之间的点积。

So, in that very special case, x would be the dot product of the first column with the output vector, and y would be the dot product of the second column with the output vector.

因此，在那个非常特殊的情况下，x 会是第一列与输出量的点积， y 会是第二列与输出量的点积。

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的点乘，所以使用这个公式。

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

这个神秘的输入量的纵坐标就是它和第一个基量(0)的点积。

Alright, so look back at this conservation of momentum expression, telling us that the dot product between this square-roots-of-the-masses vector with our change vector is the same before and after the collision.

好的， 回顾一下这个动量守恒的表达式，告诉质量平方根量与变化量的点积在碰撞前后是相同的。

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

与其使用点乘方程，让使用不带点乘的电通量方程。

This is why taking the dot product with a univector can beinterpreted as projecting a vector onto the span of that unit vector and taking the length.

这就是为什么与一个单位量做点积可以解释为将一个量投影到这个单位量张成的空间上 然后求长度。

Likewise, if things start off perpendicular, with dot product zero, like the two basis vectors, there's no guarantee that they will stay perpendicular after the transformation, preserving that zero dot product.

同样， 如果一开始两个量是垂直的，点积为零， 就像两个基量那样，不能保证它在变换后仍然保持垂直，即点积仍为零。

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， 需要实现B和ds的点积，与B ds cosine theta相同， 对吧！

Then after that, I'm going to give you my take on dot products, and something pretty cool that happens when you view them under the light of linear transformations.

在这之后 会给你一些关于点积的知识 还有一些很酷的东西当你在线性变换的光照下它的时候。

Some of you might like think of this as a kind of  dot product.

有些人可能会认为这类似于点积。

Luckily, this computation has a really nice geometric interpretation to think about the dot product between two vectors V and W.

幸运的是 这个计算有一个很好的几何解释来考虑两个量V和W之间的点积。