$\vec{u} \in \Bbb{R}^n$, $\vec{u}\not=\vec{0}$
$U = \{ k\vec{u}, k \in \Bbb{R} \} \subset \Bbb{R}^n$
$V=\{ \vec{v} \in \Bbb{R}^n : \vec{u}.\vec{v}=0 \} \subset \Bbb{R}^n$
$U + V = \{ \vec{u} + \vec{v}, \vec{u} \in U, \vec{v} \in V \} \subset \Bbb{R}^n$
You need to prove $\Bbb{R}^n \subset U + V$
Chose any $\vec{x}\in \Bbb{R}^n$
$(\vec{x}-\cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}).\vec{u} = \vec{x}.\vec{u} - \cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}.\vec{u} = 0$
So $(\vec{x}-\cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}).\vec{u} \in V$
So $\Bbb{R}^n \subset U + V$ and since we already have $U + V \subset \Bbb{R}^n$, $U + V = \Bbb{R}^n$
Then you need to prove $U\cap V=\{\vec{0}\}$
Chose any $\vec{x}\in U\cap V$
$\exists k\in\Bbb{R},\vec{x} = k \vec{u}$ and $0 = \vec{u}.\vec{x} = \vec{u}.k\vec{u} = k\|u\|^2$
So $0 = k$ because $\vec{u}\not= \vec{0}$
$\vec{x} = k \vec{u} = 0 \vec{u} = \vec{0}$
You know that $n = dim(\Bbb{R}^n) = dim(U+V) = dim( U ) + dim( V )- dim( U \cap V ) = 1 + dim( V ) - 0$
$n = 1 + dim( V )$
$dim( V ) = n - 1$