How to show two points in $\mathbb{R}^3$ form a plane and determine equation? Given two arbitrary equidistant points in $\mathbb{R}^3$, ($p$ and $q$), how would one show that they form a plane and what would the equation of that plane be?
Defining two vectors in $\mathbb{R}^3$: $(p - 0)$ and $(q - 0)$, the span of these two vectors show form a plane, if I'm not mistaken.  But now, how can I come up with an equation for this plane?
 A: The cross product of the two vectors $\vec{OP}$ and $\vec{OQ}$ is a vector perpendicular to your plane. Say it has coordinates $\vec{OP}\times \vec{OQ}=(a,b,c)$. Then the equation of the plane you are interested is $ax+by+cz=0$. This is because this represents the equation $(a,b,c)\cdot (x,y,z)=0$ which describes the set of all points whose position vector is perpendicular to $(a,b,c)$.
A: In general, a plane has an equation of the form $$ax+by+cz+d=0,$$ and you are trying to find the values of the coefficients, $a$, $b$, $c$, and $d$. You want the plane to go through the origin, $(0,0,0)$; plugging that point in, we get $d=0$, so now we have $$ax+by+cz=0.$$ Now you want your plane to go through ${\bf p}=(p_1,p_2,p_3)$ and ${\bf q}=(q_1,q_2,q_3)$, so that gives you two equations, $$ap_1+bp_2+cp_3=0,\qquad aq_1+bq_2+cq_3=0$$ to solve for the three unknowns $a$, $b$, $c$. Can you find a solution to a pair of linear equations in three unknowns?
A: It takes three non-colinear points to determine a plane.  It sounds like your plane is determined by the origin as well as $p$ and $q$.  What happens if $p=-q$?
A plane through the origin is determined by a normal vector.  How about $p \times q$?
