Home  18.013A  Chapter 3  Section 3.3 


Prove that the dot product is invariant under rotation of coordinates.
Solution:
Because the dot product is linear in each of its arguments, if we prove this statement for dot products of basis vectors, it will be true for any sum of them and hence for all vectors.
By symmetry, we need only prove that
$(\widehat{i},\widehat{i})$
which is 1, is the dot product of the image of
$\widehat{i}$
with itself under rotation of coordinates, and that
$(\widehat{i},\widehat{j})$
, which is 0, is the dot product of the image of
$\widehat{i}$
with that of the image of
$\widehat{j}$
. The same thing will be true, by symmetry with any other choice of basis vectors.
The image of
$\widehat{i}$
under rotation in the
$\widehat{i},\widehat{j}$
plane is the form that
$\widehat{i}$
takes in terms of the rotated
$\widehat{i}\text{'},\widehat{j}\text{'}$
basis vectors, which is
$\widehat{i}\text{'}\mathrm{cos}\theta \widehat{j}\text{'}\mathrm{sin}\theta $
. Similarly the image of
$\widehat{j}$
is
$\widehat{j}\text{'}\mathrm{cos}\theta +\widehat{i}\text{'}\mathrm{sin}\theta $
.
Taking the dot products of these in terms of the
$\widehat{i}\text{'},\widehat{j}\text{'}$
basis, we get
${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta $
, and
$\mathrm{cos}\theta \mathrm{sin}\theta \mathrm{sin}\theta \mathrm{cos}\theta $
, which are 1 and 0 as claimed.
