NBHM 2016 linear algebra

Question

Is the statement true?

If $v$ be a vector space over $\mathbb R^5$ with usual inner product, and $W$ and $Z$ are subspaces of $V$ both with dimension 3, then there exists $z\in Z$ such that $z\ne\mathbf 0$ and $z$ is orthogonal to $W$.

Answer 1

There exists such a $z \in Z$ iff the intersection between $Z$ and the orthogonal of $W$ is non equal to zero.

The orthogonal of $W$ is of dimension $5-3 =2$. Since $Z$ is of dimension $3$, there is no reason such a $z$ have to exist (it would have been the case if the sum of the two dimensions was $> 5$).

You can think about two planes in $\mathbb R^3$, it's almost the same situation.