Plane - Linear Algebra 
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*Background Information: I am studying Linear Algebra, and I want to understand the difference between two equations regarding planes, and when exactly to use each equation in the right place.





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*Question:


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*What is the difference between $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ and $x = x_0 + tv_1 + tv_2$? 

*How should I know when to use each of them?
Syntax Clarification: $a, b,$ and $c$ are the values of the normal vector to the plane $n = (a, b, c)$, and $v_1$ and $v_2$ are the vectors that are parallel to the plane.
 A: One of those equations is a condition on $x$, $y$ and $z$ that determines whether the triple $(x,y,z)$ is on the plane. The other is an equation that tells you all the points on the plane as $t_1$ and $t_2$ vary. 
(I think you have typos in the second one: you want
$$
v = v_0 + t_1v_1 + t_2v_2.
$$
)
Here's the analog in two dimensions. You can describe the line through $(1,0)$ and $(2,1)$ either as the set of pairs $(x,y)$ that satisfy
$$
x - y -1 = 0
$$
or as the set
$$
\{ (1,0) + t(1,1) \  | \ t \in \mathbb{R}\}.
$$
You use whichever representation helps you solve the particular problem you are trying to solve.
A: As for 1, in addition to Ethan’s answer, the second equation could be realised as a condition on $\mathbf x$ too, just enclose it in “there exist scalars $t_1, t_2$ such that…”. Also, this kind of equations is usually called parametric to signify there are variables like $t_1, t_2$. (This part should be a comment, but I can’t do this yet, sorry.)


  
*How should I know when to use each of them?
  

I specifically don’t know if there’s some guidelines other than one’s own experience, but here are some things to consider:


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*The first equation is written $(\mathbf n,\mathbf x-\mathbf x_0) = 0$ coordinate-free, and it’s only available if the vector space has an inner product $(,)$. When adding dual space in consideration, this equation could be replaced with $f(\mathbf x-\mathbf x_0) = 0$ where $f\in V^*$ a vector from dual space to $V$, from which $\mathbf x$ and $\mathbf x_0$ are.

*There is a third equation for a (hyper)plane using exterior product $\wedge$: $\mathbf v_1\wedge\mathbf v_2\wedge(\mathbf x-\mathbf x_0) = 0$ where $\mathbf v_1,\mathbf v_2$ are the same as in parametric equation.

*Parametric and ‘exterior’ equations could represent any affine subspace: point, line, 2-plane, 3-plane etc., just use more $\mathbf v_i$s. Normal equation (or its modification mentioned above) could describe only hyperplanes—planes of dimension one less than dimension of the vector space. So, this is a line when we are in a plane, and as you describe it as a plane here, one could state you’re definitely studying three-dimensional space now. :) 
A: Let $M=(x,y,z)$ and $M_0=(x_0,y_0,z_0)$ 
The parametric equation of the plane is $M=M_0+\lambda \vec{u}+\mu \vec{v}$ with $(\lambda,\mu)\in\mathbb R^2$ are variable and $(\vec u,\vec v)$ a base of vectors for the plane.
We shall see it can be transformed into the Cartesian equation of the plane.
The normal vector $\vec n=\vec{u}\wedge\vec{v}=(u_yv_z-u_zv_y,u_zv_x-u_xv_z,u_xv_y-u_yv_x)=(a,b,c)$
Since $\vec n\cdot \vec u=\vec n\cdot\vec v=0$ we have 
$\overrightarrow{M_0M}\cdot\vec n=0\iff a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Reciprocally when you have the Cartesian equation, select two points $U$ and $V$ on the plane such that $\vec u=\overrightarrow{M_0U}$ and $\vec v=\overrightarrow{M_0V}$ are not colinear and we have the parametric equation.
Since the representations are interchangeable, you are free to use whichever is the more practical for the problem you have at hand. Sometimes the Cartesian equation is more suitable to solve a problem, sometimes it is the parametric equation. 
