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.

因为所选高斯面是一个与点电荷同体，电场和面积矢总是同方， 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.

由于我刚刚看到这个变化大小是恒定，点积保持不变唯一方法就是与碰撞线形成夹角保持不变。

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 通常指同一个方 点积为正 在变换过程中被拉开 以这样方式 最终得到负点积。

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

然而，我使用E dA 余弦 theta 而不是点乘。

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，点积为正， 在变换过程中被拉开，以至于点积变为负。

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

电通等于电场与dA点积积分。

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

所以当两个通常指同一个方时 点积是正。

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。

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

注意，这看起来像是两个列 [m1, m2] 和 [v1, v2] 之间点积。

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，在那个非常特殊情况下，x 会是第一列与输出点积，而 y 会是第二列与输出点积。

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

电通等于电场与面积点乘，所以我使用这个公式。

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

这个神秘输入纵坐标就是和第一个基(0)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， 回顾一下这个动守恒表达式，告诉我质平方根与变化点积在碰撞前后是相同。

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，让我使用不带点乘电通方程。

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做点积可以解释为将一个投影到这个单位张成空间上 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， 如果一开始两个是垂直，点积为零， 就像两个基那样，不能保证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， 我需要实现B和ds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 我会给你一些关于点积知识 还有一些很酷东西当你在线性变换光照下看时候。

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.

幸运是 这个计算有一个很好几何解释来考虑两个V和W之间点积。