If $\dim(A) \lt \dim(B)$, show that there is a non-null vector of $B$ orthogonal to every vector of $A$
Let $A$ and $B$ be subspaces of a finite dimensional vector space $V$ with inner product. If $\dim(A) \lt \dim(B)$, show that there is a non-null vector of $B$ orthogonal to every vector of $A$.
Let $\{a_1,...,a_m\}$ be a basis of $A$. For every $b\in B$ consider $L(b)=((b,a_1),...,(b,a_m))$, $L(b)$ is an $m$-dimensional vector. The map $L$ is a linear transformation from $B$ to an $m$-dim space. Since $dim(B)>m$, $L$ has a nontrivial kernel. Let $0\ne b\in B$ be such that $L(b)=(0,0,...,0)$. Then $b$ is orthogonal to $A$.