Are j and k on different imaginary planes than i?

That is correct, $i$, $j$ and $k$ are contained in three separate complex planes contained in the quaternion numbers.

Just as the complex numbers $\mathbb{C}$ can be thought of as a 2-dimensional vector space over $\mathbb{R}$ with basis $1,i$, the quaternions $\mathbb{H}$ are a 4-dimensional vector space over $\mathbb{R}$ with basis $1,i,j,k$.

In particular, the plane spanned by $1,i$, the plane spanned by $1,j$, and the plane spanned by $1,k$ are different subplanes, any two of which intersect each other in the real line spanned by $1$.

So you can indeed think of these three planes as three separate copies of the complex numbers embedded in the quaternion numbers.

One way to view quaternions is using matrices: $$ 1=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} $$ $$ i=\begin{bmatrix} 0&1&0&0\\ -1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{bmatrix} $$ $$ j=\begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0 \end{bmatrix} $$ $$ k=\begin{bmatrix} 0&0&0&1\\ 0&0&-1&0\\ 0&1&0&0\\ -1&0&0&0 \end{bmatrix} $$ We can view this as $4$ orthogonal vectors which span a $4$ dimensional subspace of $\mathbb{R}^{16}$.

So yes, these can be viewed as $4$ orthogonal basis vectors, $3$ of which are orthogonal to the reals, thus "imaginary".