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?

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Sorry... how could two points not be equidistant (from each other)?!?!? – The Chaz 2.0 Oct 16 '11 at 3:11
"how would one show that they form a plane and what would the equation of that plane be?" It's not clear what you mean by two points forming a plane, but I can imagine some possibilities. (1.) The two vectors $OP$ and $OQ$ form a plane, where $P$ and $Q$ are the points with position vectors $p$ and $q$, and $O$ is the origin (assuming neither $P$ nor $Q$ is the same as $O$). This is the plane containing the three points $O$, $P$ and $Q$. (2.) The set of points $A$ equidistant from both $P$ and $Q$ is a plane perpendicular to the line $PQ$ and bisects the line segment $PQ$. – Srivatsan Oct 16 '11 at 3:11
Equidistant to what? You know you always need three conditions to uniquely determine a plane... – J. M. Oct 16 '11 at 3:11
Okay, you're talking about three points: $p$ and $q$ plus the origin (in which case $\|p\|=\|q\|$ isn't really relevant). Do you know what a cross product is? – anon Oct 16 '11 at 3:12

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 $(a,b,c)$.

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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?

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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 $p \times q$?

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