Calculate the vector normal to the plane by given points How can one calculate the vector normal to the plane that is determined by given points?
For example, given three points $P_1(5,0,0)$, $P_2(0,0,5)$ and $P_3(10,0,5)$, calculate the vector normal to the plane containing these three points.
The compute the normal is by vector product.
$$
a = \left(\begin{matrix} x_2-x_1\\y_2-y_1\\z_2-z_1 \end{matrix} \right)
\qquad
b = \left(\begin{matrix} x_3-x_1\\y_3-y_1\\z_3-z_1 \end{matrix} \right)
$$
therefore $a = -5i+5k$, and $b=5i+5k$
$$
a\times b = \left|\begin{matrix}
    i & j & k \\
    -5 & 0 & 5 \\
    5 & 0 & -5 \\
\end{matrix}\right|
$$
The questions are:
1) Are A and B is given, that means no matter which three point i use, the $a$ and $b$ is still using this to calucate?
2) How to get $a$ , $b$ and $a\times b$ ? Thank you
 A: Given any two vectors $a$ and $b$, the vector cross product $a\times b$ will be perpendicular to both of them, even if we have to consider the degenerate case of $0$ (that is, zero is perpendicular to anything).
With three distinct points $x_1,x_2,x_3$ we have a triangle, and this triangle determines a plane. Note that the triangle is determined by any two of its sides - and so is the plane then. Sides correspond to displacement vectors between the three points, i.e. we can set $a=x_2-x_1$, $b=x_3-x_1$ without losing generality (that is, it won't matter what order we understand the points to be in, the resulting displacement vectors will determine the same exact plane). Finally, since the points are distinct, the displacement vectors are nonzero, and if we assume they are noncollinear the displacements are also nonparallel, so their cross product will be nondegenerate (nonzero). (Non non non non...)
Since we're working in three dimensions, any vector perpendicular to these two displacement vectors will be perpendicular to the plane they determine, and vice versa, so we can use the cross product of these displacement vectors to determine the normal vector, up to various scalings.
With $i,j,k$ the orthonormal basis, the cross product is computed via the "pseudo-determinant"
$$a\times b = \det \begin{bmatrix}i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix}=(a_2b_3-a_3b_2)i-(a_1b_3-a_3b_1)j+(a_1b_2-a_2b_1)k.$$
Where $a=a_1i+a_2j+a_3k$ and $b=b_1i+b_2j+b_3k$.
