This is a step of a proof in the book Variational Problems in Geometry by Seiki Nishikawa. I will ignore the background and change some of the statements and notations for simplicity.

Let $(M,g)$ be a smooth Riemannian manifold. Suppose $w:M\to\mathbb R^q$ is an isometric embedding. Let $N$ be a tubular neighborhood of $w(M)$, and $\pi:N\to w(M)$ the canonical projection. Let $\tau(w)$ denote the tension field of $w$. Then since $w$ is an isometric embedding, $\tau(w)$ is orthogonal to $w(M)$ and hence $d\pi(\tau(w))=0$.

What I don't understand is the emphrasized sentence. Why does $w$ being isometric imply $\tau(w)\perp w(M)$?