When we dot product normal vector and a vector on the plane, how does it give us the equation of the plane. I mean, how can we even say vector "on" the plane since the vector is just a length and direction?
I understand that the dot product will be zero, but if I take these two vectors anywhere in the 3d space, shouldn't the answer be the same? If so then how is that the equation of the plane?
The line $a$ defined perpendicular to the plain $\pi$ if for all line $b\subset\pi$ we have $a\perp b$.
Now, let $\vec{n}(A,B,C)$ be a normal for the plain $\pi$ and $M(x,y,z)\in\pi$, $A(x_1,y_1,z_1)\in\pi$.
Thus, for all $M\in\pi$ we obtain: $$\vec{AM}\perp\vec{n}$$ or $$(x-x_1,y-y_1,z-z_1)(A,B,C)=0$$ or $$A(x-x_1)+B(y-y_1)+C(z-z_1)=0.$$ Now, let $-Ax_1-By_1-Cz_1=D$.
Thus, we got an equation of the plain: $$Ax+By+Cz+D=0.$$
Two arbitrary orthogonal vectors in $\mathbb{R}^{3}$ need not determine a pre-specified plane. It is that if a point $(a,b,c)$ on a given plane $E \subset \mathbb{R}^{3}$ is given and if a normal vector $(l,p,q)$ to the plane is given, then $(x,y,z) \in E$ if and only if $(x-a,y-b,z-c)^{\top}(l,p,q) = 0$. The equation is called Cartesian equation of $E$.
Note that in Euclidean geometry it is concentional to view two parallel vectors of the same length as the same vector. Note that the dilation of a vector is not supposed to change its direction.
I am going to focus on this question: how can we even say vector "on" the plane?
Short answer: we can't.
If you recall the algebraic properties of vector spaces, vectors can be multiplied by constants. For instance, if you consider a line as a vector space, it must cross the origin. So to speak, all vectors are pinned at the origin. We need an additional concept to represent arbitrary lines and planes in 3D space.
In affine geometry, we consider points. An affine space is a vectorial space plus a point. So you are allowed to say that a point is on a plane (a bidimensional affine space). Without entering into much rigour, the difference between two points is a vector $v=p_2-p_1$, or you can add a vector to a point to reach another point $p_2=p_1+v$.
In your example, you provide a vectorial space $V$ in any way, e.g. giving a normal vector; and you also consider a point $p$. Now, your equation defines the plane as the set of points that can be reached from $p$ by adding vectors from $V$.
if a normal vector (l,p,q) to the plane is given, then (x,y,z)∈E if and only if (x−a,y−b,z−c)⊤(l,p,q)=0.. Why? – ali9091 Aug 12 '17 at 06:13Sorry for asking again but I cannot seem to get it. Your help is greatly appreciated.
– ali9091 Aug 12 '17 at 15:50