Hyperplanes and intersection How to prove that if I have two hyperplanes in $\mathbb{R}^{n}$ that have only one point of intersection, then $n=2$  (or $n=1$ trivially).
 A: Hint
A hyperplane is parameterised by
$$H(x,d) := \{y \in \mathbb{R}^n | \langle x,y \rangle = d\}$$
Show that the Intersecion of two such planes is at least $(n-2)$-dimensional if nonempty.
A: Without loss of generality, we may assume that the origin is a point of intersection. 
A hyperplane is given by a single linear equation, i.e.
$$a_1x_1 + a_2x_2 + \cdots + a_nx_n = 0$$
where each of the $a_i$ are real numbers and not all of them are zero. If you have a second hyperplane:
$$b_1x_1 + b_2x_2 + \cdots + b_nx_n = 0$$
where each of the $b_j$ are real numbers and not all of them are zero.
The intersection is given by the set of points on both planes, i.e. the set of $(x_1,x_2,\ldots,x_n)$ with
\begin{array}{ccc}a_1x_1 + a_2x_2 + \cdots + a_nx_n &=& 0 \\
b_1x_1 + b_2x_2 + \cdots + b_nx_n &=& 0
\end{array}
If the vectors $(a_1,a_2,\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ are linearly independent then you have two independent linear equations is $n$ unknows. You can solve the first equation for one of the $x_i$, then substitute this into the second equation and solve for one of the $x_j$, where $i \neq j$. We have solved for two of the unknowns, leaving $n-2$ free unknows. In other words, if the two planes are not coincident, their intersection will be a linear subspace of dimension $n-2$. If you want this to be a point, which has dimension zero, then you want $n-2=0$, i.e. $n=2$.
