translation and rotation in Euclidean space

Clearly taking a vector and rotating it, then translating it, does
nothing to the distance. So, under translation and rotations, the
Euclidean standard metric is invariant. Am seeking to show this in the
plane. clearly,
(x,y,1)dot(x,y,1)=x^2+y^2+1 Now I wish to translate and rotate a
vector, and check that the distance is the same.

(a -b 0) (1 0 u)
(b a 0) (0 1 v)
(0 0 1) (0 0 1) are the rotation and translation
matrices, the result of multiplying them is

(a -b au-bv )
(b a bu+av)
(0 0 1 ) now to apply them to a vector (x,y,1) we
get


(ax-by+au-bv )
(bx+ay+bu+av )
( 1 ) now we dot this vector with itself and
do some simplification to the resulting scalar...



(a^2+b^2)[ (x+u)^2 + (y+v)^2)] +1 now (a^2+b^2)=1 so we have (x
+u)^2 +(y+v)^2 + 1

the question, am supposed to get a scalar equal to x^2 + y^2 +1.
Clearly we have only translated
the vector, but I must be getting a definition wrong, somewhere mixing
scalars and vectors in a wrong way.

Is this correct, to translate a vector,


(1 0 u) (x) (x+u)
(0 1 v) x (y) = (y+v)
(0 0 1) (1) ( 1 ) but to me, this vector isn't the vector
(x,y,1) translated, it's the addition
of two vectors, and will have a different, longer distance. Thanks in
advance for any help.
Best...Frank


.

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