Prove that the sum of 2 bivectors in $R^{3}$ is a bivector? Hint:Think geometrically $R^{3}$ is noted as a 3 dimensional space. I know that pseudo-scalars are bivectors on a plane.  also I know bivectors have more freedom than scalars and vectors on a plane.  Can I use $a \land b$ + $c \land d$ to start off this proof?
 A: There is some redundancy in expressing bivectors with the symbols $a\wedge b$. It is possible that $a\wedge b=c\wedge d$ even if $a,b$ are different from $c,d$ (including up to sign). We can think of any bivector $a\wedge b$ as representing a 2D subspace with orientation determined by the basis $\{a,b\}$, but equipped with a scalar equal to the area of the parallelogram spanned by $a$ and $b$. Therefore, if $R$ is a rotation that stabilizes the plane spanned by $a$ and $b$, the rotation preserves the area of the parallelogram spanned by $a$ and $b$, so $a\wedge b=(Ra)\wedge (Rb)$.
(If we assume $a,b$ are L.I. and use a basis $\{a,b,\bullet\}$ in order to write an explicit formula for $R$, this fact can be checked explicitly with algebra.)
Therefore, to simplify a sum $a\wedge b+c\wedge d$ to a simple bivector, we note that the 2D subspaces spanned by $\{a,b\}$ and $\{c,d\}$ intersect in some axis $\ell$, then apply rotations $R$ and $S$ to $a,b$ and $c,d$ respectiely so that it looks like $p\wedge q+r\wedge s$ with $p\| r$ parallel. When they're parallel, they can be expressed as scalar multiples of a common vector, and then you can apply the distributive property for $\wedge$ only in reverse.
