The reason behind the name "Orthogonal transformation". An orthogonal transformation is a linear transformation such that $(Tx,Ty)=(x,y)$. 
Orthogonality is suggestive of perpendicularity. What might have been the reason for naming a distance preserving linear transformation on a vector space "orthogonal"?
Thanks!
 A: I think it's due to the fact that, in a finite dimensional vector space, the columns of the associated matrix of the transformation are mutually orthogonal.
They don't just preserve distance, they also preserve inner product. In more familiar examples, you can think of this as preserving the angle between two vectors. 
A: Over a finite dimensional vector space, the following are equivalent:


*

*$T$ preserves inner products (i.e. angles): $\langle Tx,Ty\rangle=\langle x,y\rangle$ for each $x,y$.

*$T$ preserves orthonormal bases: For every orthonormal basis $B$, $T(B)$ is an orthonormal basis.

*$T$ preserves an orthonormal basis: There exists an orthonormal basis $B$ such that $T(B)$ is an orthonormal basis.

*$T$ preserves distances $\lVert Tx\rVert =\lVert x\rVert$ for each $x$.

*$TT^\ast=T^\ast T=1$


They all define what we call an orthogonal transformation, one that preserves all geometrical relations in our space.
A: I have insufficient reputation to add a comment, so I'll expand Carsten's comment into a response.
Like you say, an orthogonal transformation satisfies $(Tx, Ty) = (x,y)$.  Let's look at some simple consequences.  First off, if $x\perp y$, so that $(x,y) = 0$,
then $Tx\perp Ty$.  This is one reason $T$ might be called orthogonal: it preserves orthogonality of vectors.
Another thing we can do is look at the dual map $:T^*:V\to V$, which satisfies 
$(Tx, y) = (x, T^*y)$.  (Here I've identified $V$ with its dual, since we have defined an inner product on $V$).  If $V$ is a finite-dimensional space over $\mathbb{R}$, and $(x, y) = y^T x$, and $T$ is represented by multiplication by a matrix, then $T^*$ is multiplication by its transpose.  To see this, we just
compute
$$
(Tx, y) = y^T Tx = (y^T Tx)^T = x^T T^T y = (x, T^*y).
$$
Now if we use the orthogonality condition, we have $(Tx, Ty) = (x, T^*Ty) = (x,y)$.  Thus, $T^*T = Id$ is the identity operation.  In terms of matrices,
this means that the rows of $T$ must be orthogonal to each other.  Similarly,
since $TT^* = Id$, the columns of $T$ must also be orthogonal.  This is another sense in which $T$ is orthogonal.
A: The matrix form/representation for an orthogonal transformation is an orthogonal matrix, which has the property that the row vectors are orthogonal to each other. That explains the "perpendicularity" part of your question.
