Prove that if <a>x<b> + <b>x<c> + <c>x<a> = 0 then vectors <a>, <b>, and <c> are coplanar.
Note: <u> here means vector u.
Claim: If three vectors add to zero, they must be coplanar.
Proof: For three vectors to add to zero, you must start at the initial point of one, travel
along it, then travel along the second vector, and then travel along the third and end up where
you started. This route forms a triangle (or a degenerate triangle), and since there is no such thing
as a three dimensional triangle, the three vectors must be coplanar.
Implies: <a>x<b>, <b>x<c>, and <c>x<a> are coplanar.
<a> and <b> are coplanar, because any two vectors are.
If <c> not being coplanar with <a> and <b> implies that <a>x<b>, <b>x<c>, and <c>x<a> are
not coplanar, then the conjecture is proved by contradiction.
Assume: <c> is not in the same plane as <a> and <b> and show that this assumption would mean
that <a>x<b>, <b>x<c>, and <c>x<a> are not in the same plane.
<a>x<b> and <b>x<c> are in the same plane because any two vectors are coplanar.
Showing that <c> not being coplanar to <a> and <b> means that <c>x<a> is not coplanar to
<a>x<b> and <b>x<c> proves the conjecture.
Consider a plane full of vectors, and each of these vectors is a normal line to another plane. Call
these planes beta planes, and call the original plane (the one full of vectors) the alpha plane. Beta
planes intersect other beta planes and the alpha plane, but they can never intersect another beta
plane in such a way that the resulting line would not be normal to the alpha plane. This comes
from the definition of the normal line. Since the alpha plane contains normal lines to all the beta
planes, each beta plane has to be perpendicular to its given normal line, and by extension the
alpha plane which contains it. This means that lines formed at the intersections of beta planes
must be perpendicular to the alpha plane as well.
The plane containing <a>x<b> and <b>x<c> is our alpha plane. This makes the plane containing
<a> and <b> and the plane containing <b> and <c> both beta planes. If <c>x<a> is indeed
coplanar to <a>x<b> and <b>x<c>, then <c>x<a> is also in the alpha plane and the plane
containing <c> and <a> is also a beta plane. If then, we can show that <c> not being coplanar to
<a> and <b> means that not all of the three proposed beta planes can be beta planes or that not
all of the three cross products can be in the alpha plane, or that any of the proposed beta planes
intersect in such a way that the resulting line is not normal to the proposed alpha plane, then the
conjecture is proved.
For convenience, set the beta plane containing vectors <a> and <b> as the xy plane. According
to the assumption we made in attempt to prove the conjecture by contradiction, <c> is not in this
beta plane, though we can put its initial point at the origin, along with the initial points of vectors
<a> and <b>, again, for convenience. Now, the lines formed when beta planes intersect each
other, (call them gamma lines), are all perpendicular to a common alpha plane, but this implies
that all the gamma lines are parallel to each other. Now, the beta plane containing <a> and <b>
intersects the beta plane containing <b> and <c> intersects at gamma line <b>. The beta plane
containing <a> and <c> intersects the beta plane containing <c> and <b> at gamma line <c>.
The beta plane containing <b> and <a> intersects the beta plane containing <c> and <a> at
gamma line <a>. Now, since <a>, <b>, and <c>, share a common point, the origin, they cannot
be parallel, which means that not all of them can be normal to the alpha plane, which means a
common alpha plane cannot exist, which means that <a>x<b>, <b>x<c>, and <a>x<c> cannot be
coplanar, which we established early on is impossible if they sum to zero. Therefore, if <a>x<b>
+ <b>x<c> + <c>x<a> = 0, then <a>, <b>, and <c> are coplanar.