Let $K$ be a field of characteristic different from two, and $Q\subset\mathbb{P}^{n+1}_K$ an $n$-dimensional smooth quadric hypersurface over $K$.
Suppose also that $Q$ has a $K$-point and so $Q$ is rational over $K$. Is there any result about the existence of a $k$-plane $H\cong\mathbb{P}^k\subset Q$ also defined over $K$ for $n$ big enough? And if so can we determine what is the smallest value of $n$ (as a function of $k$) for which $H$ exists?
Let $P \in Q$ be a $K$-point. First, if $Q$ has a $k$-plane not passing through $P$, it also has a plane passing through $P$. Indeed, if $P \not\in H \subset Q$, consider the linear span $\tilde{H}$ of $P$ and $H$. If $\tilde{H} \not\subset Q$ then $Q \cap \tilde{H}$ is the union of two $k$-planes --- $H$ and the residual $k$-plane $H'$ --- and since $P \in (Q \cap \tilde{H}) \setminus H$, we have $P \in H' \subset Q$. Alternatively, we can take $H'$ to be any $k$-plane in $\tilde{H}$ containing $P$.
So, the question reduces to the question about $k$-planes contaning $P$. Now consider the hyperbolic reduction of $Q$ with respect to $P$ --- if $Q$ is defined by a quadratic form $q$ on a vector space $V$ and the point $P$ corresponds to a vector $v \in V$, consider the space $\bar{V} := v^\perp / v$ (the orthogonal is taken in $V$ with respect to $q$ and $v \in v^\perp$ because $P \in Q$). The form $q$ induces a non-degenerate quadratic form $\bar{q}$ on $\bar{V}$, and the hyperbolic reduction of $Q$ is defined as the corresponding quadric $\bar{Q} \subset \mathbb{P}(\bar{V})$.
Note also that the hyperbolic reduction procedure is invertible --- given a quadric $\bar{Q} \subset \mathbb{P}(\bar{V})$ defined by a quadratic form $\bar{q}$ one can take the quadric $Q$ corresponding to the sum of $\bar{q}$ with a hyperbolic form and a point $P$ on the hyperbolic summand of $Q$; then the hyperbolic reduction of $Q$ with respect to $P$ gives back $\bar{Q}$.
Finally, it is easy to see that there is a bijection between the set of $k$-planes through $P$ in $Q$ and the set of $(k-1)$-planes in $\bar{Q}$. So, a priori the existence of a point on $Q$ does not imply anything about the existence of $k$-planes with $k > 0$.
