Need help determining the pairs of quaternions that anticommute I tried to solve another exercise and I would be grateful if someone could tell me if my answer is right. This is the exercise:

Characterize the pairs $p,q \in \mathbb H$ such that $pq = -qp$.

I can see that the unit quaternion paris $(i,k), (i,j), (j,k)$ anticommute.
Since the real part of two purely imaginary quaternions is the dot product and the imaginary part the vector product it follows that purely imaginary quaternions anticommute.
If $a+p, b+q$ are two quaternions with real parts $a,b$ then $(a+p)(b+q) = ab + aq + bp + pq$. The only way for this to anticommute is if $ab=0$. Therefore I conclude that all pairs of purely imaginary quaternions anticommute.
Is this correct?

Claim: $q_1$ anticommutes with $q_2$ if and only if either is zero, or else both have zero real-part and their imaginary parts are orthogonal.
Representing our quaternions using a vector for the imaginary part, then the product of $q_1 = (r_1, \vec{v_1})$ and $q_2 = (r_2, \vec{v_2})$ is $(r_1r_2 - \vec{v_1}\cdot\vec{v_2}, \vec{v_1}\times\vec{v_2}+r_1\vec{v_2}+r_2\vec{v_1})$.
First, if either $\vec{v_1}$ or $\vec{v_2}$ are zero then it quickly follows that the corresponding $r_1$ or $r_2$ is also zero (or else, the other quaternion is also zero). So we can restrict ourselves to the case where both $\vec{v_1}$ and $\vec{v_2}$ are non-zero.
Considering real and imaginary parts of $q_1q_2 + q_2q_1$ (which must be zero if the quaternions anticommute):
The real part: 
$2r_1r_2 - 2\vec{v_1}\cdot\vec{v_2} = 0$ which implies $r_1r_2 = \vec{v_1}\cdot\vec{v_2}$.
The imaginary part: $\vec{v_1}\times\vec{v_2} + \vec{v_2}\times\vec{v_1} + 2r_1\vec{v_2} + 2r_2\vec{v_1} = 2r_1\vec{v_2} + 2r_2\vec{v_1} = 0$ (using the anticommutativity of vector cross product).
Dotting the last equation with $\vec{v_1}$ and substituting gives us:
$r_1\vec{v_1}\cdot\vec{v_2} + r_2\vec{v_1}\cdot\vec{v_1} = 0$
$r_1^2r_2 + r_2\|\vec{v_1}\|^2 = r_2(r_1^2 + \|\vec{v_1}\|^2) = 0$
Since we've assumed $\vec{v_1}$ is non-zero, this can only be satisfied if $r_2=0$. By symmetry we get $r_1=0$ and our conclusion that both real parts must be zero.
Finally, we can use the equation $r_1r_2 = \vec{v_1}\cdot\vec{v_2}$ to deduce that the two imaginary parts must be orthogonal.
That the condition is sufficient (that is, if both real parts are zero and the imaginary parts are orthogonal then the quaternions anticommute) can be observed from the definition of mulitplication.
