A plane goes through $A:(0,1,2), \ B:(3,2,1)$ and $C:(4,-1,0).$ Determine the equation of the plane in affin form.

Question

Solution:

The vectors $\vec{AB}=(3,2,1)-(0,1,2)=3,1,-1$ and $\vec{AC}=(4,-1,0)-(0,1,2)=(4,-2,-2),$ are two direction vectors of the plane. A normal vector $\vec{n}$ to the plane is then given by $$\vec{n}=\vec{AB}\times\vec{AC}=(-4,2,-10).$$

Since $A$ is a point on the plane, we get

$$\vec{n}\cdot (x-0,y-1,z-2)=-4x+2(y-1)-10(z-2)=-4x+2y-10z+18=0.$$

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I dont understand the first part of the last equation.

1. What vector is $(x-0,y-1,z-2)$?
2. Why do they take the dot-product of the above with the normal vector? (How does it give the equation of the plane?)

Answer 1

1. The vector $\langle x-0,y-1,z-2\rangle$ is a vector connecting a generic point $P(x,y,z)$ lying in the desired plane with the point $A(0,1,2)$, which is also in this plane. Since both of them are in this plane, the vector connecting them $\overrightarrow{AP}=\langle x-0,y-1,z-2\rangle$ is in this plane.

2. The word "normal" means "perpendicular" in this context. So saying "$\overrightarrow{n}$ is normal to the plane" means that $n$ is perpendicular to the plane, and therefore it's perpendicular to any vector that lies in this plane. So we know that $\overrightarrow{n}\perp\overrightarrow{AP}$ must be true. And then there's the property of dot products: two vectors are perpendicular if and only if their dot product is zero.

Answer 2

The vector $(x-0,y-1,z-2)$ comes directly from the fact that $(0,1,2)$ is a point on the plane. For the second part of your question, if $\vec{v}$ is a vector and $\vec{n}$ is a normal vector to $\vec{v}$, then $\vec{v}\cdot\vec{n}=0$. We use the dot product between the normal vector and the cross product of the displacement vectors because the equation of the plane is

$(x-a,y-b,z-c)\cdot \vec{n}=0$,

where $(a,b,c)$ is a point on the plane and $\vec{n}$ is normal to the plane.

Answer 3

The idea of finding the equation of a plane is to have an equation that can help to locate any point $P(x,y,z)$ on the plane. In order to do that consider the vector $\vec{AP}=(x,y,z)-(0,1,2)$. This vector lies in the given plane. Now that we have a normal vector $\hat{n}$ (which by definition is orthogonal to any vector in the plane) we know that $\hat{n} \cdot \vec{AP}=0$. This gives the equation we are looking for.