= Solution
The <Grassmann graph> is connected, and its distance can be determined directly. Put $C=A\cap B$, $r=\dim C$, and $s=k-r$. Choose <vectors> so that
$$
A=C\oplus\langle a_1,\ldots,a_s\rangle,\qquad
B=C\oplus\langle b_1,\ldots,b_s\rangle.
$$
The <vectors> of a <basis> of $C$, followed by all the $a_i$ and $b_i$, are independent: a relation would identify a <vector> in the two chosen complements modulo $C$, contradicting $A\cap B=C$. For $0\le j\le s$ define
$$
A_j=C\oplus\langle b_1,\ldots,b_j,a_{j+1},\ldots,a_s\rangle.
$$
Each successive pair has an intersection of <dimension> $k-1$. Thus $A_0=A$ and $A_s=B$ are joined by a path of length $s$.
For the lower bound, if two vertices $U,V$ are adjacent, their common <vector subspace> $D$ has codimension one in each. Both $\dim(U\cap B)$ and $\dim(V\cap B)$ lie between $\dim(D\cap B)$ and $\dim(D\cap B)+1$. Thus one edge can change the <dimension> of the intersection with $B$ by at most one. It must change from $r$ to $k$ along a path from $A$ to $B$, requiring at least $k-r$ edges. Therefore
$$
\boxed{d(A,B)=k-\dim(A\cap B).}
$$
If two ordered pairs have equal distance, their intersections have equal <dimension>. Construct adapted <bases> for each pair as above and extend both to <bases> of the ambient <vector space>. The invertible <linear map> between them sends the first ordered pair to the second. It preserves intersection <dimensions>, so it is a <graph automorphism>. This proves that the <Grassmann graph> is <distance-transitive>, with the required simultaneous images. Its diameter is $k$, because $2k\le n$ permits two disjoint $k$-dimensional <vector subspaces>.
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