# Normal space and tangent space

thanks for the help the basis of R^3 must be 3 vectors . I am not understand why v in R^3 is written as summing two vectors , one from Normal Space and the other from tangent space

• Where is that definition from? It looks as if the normal space would be merely a line in any case...which is true in $\;\Bbb R^3\;$ . The definition I know is that the normal space to $\;M\;$ at point $\;p\;$ is the orthogonal complement to the tangent space (or plane) to $\;M\;$ at $\;p\;$ ... – DonAntonio May 19 '17 at 13:49

Since $\mathbb{R}^3$ has dimension three, any basis for $\mathbb{R}^3$ has three elements. $M$ is a surface, i.e. a $2$-manifold on $\mathbb{R}^3$, and so if $p\in M$ then the tangent space $T_pM$ is a vector space of dimension $2$. Thus, let $v_1, v_2$ be a basis for $T_pM$. Now let $v_3 \in N_pM$ be nonzero. Then the set $\{v_1, v_2, v_3\}$ is a basis for $\mathbb{R}^3$, since it is linearly independent and has three vectors. Now let $x\in \mathbb{R}^3$ be arbitrary. Then we may write $$x = c_1 v_1+c_2v_2+c_3v_3$$ for some scalars $c_1, c_2, c_3$. $v_1, v_2 \in T_pM$, and so $v_t:=c_1v_1+c_2v_2 \in T_pM$. $v_3\in N_pM$, and so $v_n := c_3v_3 \in N_pM$. Therefore, we have $$x = v_t+v_n$$ where $v_t \in T_pM$, and $v_n \in N_pM$, as your book claims. However, $\{v_t, v_n\}$ is not a basis for $\mathbb{R}^3$; for a different choice of $x$, you'll need completely different values for these vectors.
As you noted, $\mathbb R^3$ is the direct sum of two vector spaces, namely the tangent space and the normal space. Now look up the definition of a direct sum, please.