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 $\tau(w)\perp w(M)$?

**Edit:**
The tension field $\tau(w)$ is an $\mathbb R^q$-valued vector field on $M$, or more precisely, a smooth section of the bundle $w^{-1}(T\mathbb R^q)$. It is defined in coordinates ($x^i$ on $M$ and the standard coordinates on $\mathbb R^q$) by $\tau(w)^r=\Delta w^r+g^{ij}\hat\Gamma_{kl}^rw^k_iw_j^l$, where $g$ is the metric on $M$ and $\hat\Gamma_{ij}^r$ are the Christoffel symbols of $\mathbb R^q$. But since w.r.t. the standard coordinates of the Euclidean space all coefficients of its metric are $0$, $\hat\Gamma_{ij}^r=0$ and $\tau(w)=\Delta w$.