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I'm reading this: on vectors

and basically it seems like the dot product of a normal vector to a plane and a vector on the plane is equal to the equation of the plane. What is the intuition behind this?

It seems like the equation of the plane is Ax + By + Cz = D

I was watching this video and it seems like you can define a plane just with a normal vector and a point on the plane. But how do you know how big this plane is?

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  • $\begingroup$ What do you mean by how big the plane is? Have you studied dot products? $\endgroup$
    – Karl
    Jan 27, 2016 at 18:53
  • $\begingroup$ The dot product of normal to the plane and any other vector is equal to the normal distance of the plane from the origin: if and only if the vector describes a point lying on that plane. $\endgroup$
    – ARi
    Jan 27, 2016 at 18:55
  • $\begingroup$ Have you seen $a \cdot b = |a||b| \cos \theta$ ? $\endgroup$
    – Karl
    Jan 27, 2016 at 18:58
  • $\begingroup$ yes I've seen that equation. It's law of cosines right? $\endgroup$
    – Jwan622
    Jan 27, 2016 at 19:16
  • $\begingroup$ The dot product is a way of measuring how perpendicular the vectors are. $\cos 90^{\circ} = 0$ forces the dot product to be zero. Ignoring the cases where the magnitude of the vectors is zero anyway. $\endgroup$
    – Karl
    Jan 27, 2016 at 19:34

2 Answers 2

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The linked reading isn't saying that the dot product is equal to the equation of the plane, it's saying that setting the dot product equal to 0 gives the equation of the plane. Following the notation of the linked page, let $\vec{n} = \langle a, b, c \rangle$ be the vector normal to the plane, let $\vec{r}_{0}$ be the position vector of a point in the plane $P_0 = (x_0, y_0, z_0)$, and let $\vec{r}$ be the position vector of an arbitrary point in the plane $P = (x, y, z)$. $a, b, c, x_0, y_0, z_0$ are all known; $x, y, z$ are free variables. The vector $\vec{r} - \vec{r}_{0}$ is in the plane, implying it's orthogonal to $\vec{n}$, thus $\vec{n} \cdot (\vec{r} - \vec{r}_{0}) = 0$. Substituting the component forms of the vectors into this equation gives us an equation in $x, y, z$ that defines the plane.

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  • $\begingroup$ The plane extends infinitely in all directions right? $\endgroup$
    – Jwan622
    Jan 27, 2016 at 19:18
  • $\begingroup$ @Jwan622 Yes, by definition. $\endgroup$
    – DylanSp
    Jan 27, 2016 at 19:31
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The product of a normal vector and a vector on the plane gives 0. This forms an equation we can use to get all values of the position vectors on the plane when we set the points of the vectors on the plane to variables x, y, and z. So when we plug in the given values into our dot product equals to 0 equation with the x, y, and z as unknown variables, we get an equation where we have everything known except the x, y, and z points on the plane and therefore the equation of the plane itself. That's the intuition behind the Khan Academy video you linked.

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