Theorem: Vector Magnitude is Invariant Under Rotation

Let $\mathbf v$ be an arbitrary vector in the Cartesian plane $\CC$.
Let the coordinate system then be rotated in the anticlockwise direction by an arbitrary angle $\theta$.
Then:
the magnitude of $\mathbf v$ is unchanged in the new coordinate system.

Proof:
Let $P = \tuple {x_1, y_1}$ be the initial point of $\mathbf v$.
Let $Q = \tuple {x_2, y_2}$ be the terminal point of $\mathbf v$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf v
      | r = \tuple {X, Y}
      | c = {{Defof|Vector Quantity}}
}}
{{eqn | l = \tuple {X, Y}
      | r = \tuple {x_2 - x_1, y_2 - y_1}
      | c = {{Defof|Vector Quantity}}
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = \size {\mathbf v}
      | r = \sqrt {X^2 + Y^2}
      | c = {{Defof|Vector Length}}
}}
{{eqn | r = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}
      | c = Substitution
}}
{{end-eqn}}
where $\size {\mathbf v}$ denotes the magnitude of $\mathbf v$.
The square of $\size {\mathbf v}$ is:
{{begin-eqn}}
{{eqn | l = \size {\mathbf v}^2
      | r = \paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2
}}
{{eqn | r = {x_2}^2 - 2 x_1 x_2 + {x_1}^2 + {y_2}^2 - 2 y_1 y_2 + {y_1}^2 
}}
{{end-eqn}}
By Equations defining Plane Rotation:
:$x' = x \cos \theta - y \sin \theta$
:$y' = x \sin \theta + y \cos \theta$
Let the vector in the rotated coordinates be $\mathbf v'$.
The $x$-component of $\mathbf v'$ is:
:$X' = x_2 \cos \theta - y_2 \sin \theta - x_1 \cos \theta + y_1 \sin \theta$
The $y$-component of $\mathbf v'$ is:
:$Y' = x_2 \sin \theta + y_2 \cos \theta - x_1 \sin \theta - y_1 \cos \theta$
We compute the square of $X'$ and $Y'$ separately:
{{begin-eqn}}
{{eqn | l = {X'}^2
      | r = \paren {x_2 \cos \theta - y_2 \sin \theta - x_1 \cos \theta + y_1 \sin \theta}^2
}}
{{end-eqn}}
There are $16$ terms so we group them on four lines as a bookkeeping device:
{{begin-eqn}}
{{eqn | l = {X'}^2
      | r = {x_2}^2 \cos^2 \theta + {y_2}^2 \sin^2 \theta + {x_1}^2 \cos^2 \theta + {y_1}^2 \sin^2 \theta
}}
{{eqn | o = -
      | r = 2 x_2 y_2 \sin \theta \cos \theta - 2 x_1 x_2 \cos^2 \theta + 2 x_2 y_1 \sin \theta \cos \theta 
}}
{{eqn | o = +
      | r = 2 x_1 y_2 \sin \theta \cos \theta - 2 y_1 y_2 \sin^2 \theta 
}}
{{eqn | o = -
      | r = 2 x_1 y_1 \sin \theta \cos \theta
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = {Y'}^2
      | r = \paren {x_2 \sin \theta + y_2 \cos \theta - x_1 \sin \theta - y_1 \cos \theta }^2
}}
{{end-eqn}}
Group the terms on separate lines:
{{begin-eqn}}
{{eqn | l = {Y'}^2
      | r = {x_2}^2 \sin^2 \theta + {y_2}^2 \cos^2 \theta + {x_1}^2 \sin^2 \theta + {y_1}^2 \cos^2 \theta 
}}
{{eqn | o = +
      | r = 2 x_2 y_2 \sin \theta \cos \theta - 2 x_1 x_2 \sin^2 \theta - 2 x_2 y_1 \sin \theta \cos \theta 
}}
{{eqn | o = -
      | r = 2 x_1 y_2 \sin \theta \cos \theta - 2 y_1 y_2 \cos^2 \theta 
}}
{{eqn | o = +
      | r = 2 x_1 y_1 \sin \theta \cos \theta 
}}
{{end-eqn}}
All the terms with $\sin \theta \cos \theta$ cancel.
We have six pairs of terms remaining.
By Sum of Squares of Sine and Cosine:
:$\sin^2 \theta + \cos^2 \theta = 1$
So:
{{begin-eqn}}
{{eqn | l = \size {\mathbf v'}^2
      | r = {x_2}^2 - 2 x_1 x_2 + {x_1}^2 + {y_2}^2 - 2 y_1 y_2 + {y_1}^2
}}
{{eqn | r = \paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2
}}
{{end-eqn}}
Compare with the results for $\mathbf v$:
:$\size {\mathbf v'}^2 = \size {\mathbf v}^2$
Since magnitude is positive, when we take the square root:
:$\size {\mathbf v'} = \size {\mathbf v}$
{{qed}}