Solution (#900) Let a, b be vectors in R3 .
If a = 0 then the equation is equivalent to r · b = 0 which is the equation of a plane which, in particular, means
that there are non-zero solutions.
Say now that a = 0. Our original equation is r ∧ a = (r · b)r and so if we dot with a we see
(r · b)(r · a) = 0.
So either r · a = 0 or r · b = 0.
Case (i): r · a = 0. This then means that
(r ∧ a) ∧ a = (r · a)a − |a|2 r = − |a|2 r.
Thus we have
− |a|2 r = ((r · b)r) ∧ a = (r · b)2 r.
2
If r = 0 then it also follows that (r · b)2 + |a| = 0, a contradiction.
Case (ii): r · b = 0. We then have r ∧ a = 0 and so r = λa for some λ = 0 (if r is to be a non-zero solution). Hence
we can only have non-zero solutions in this case if a · b = 0.
In summary then, there are non-zero solutions r when a · b = 0.