Vector Magnitude is Invariant Under Rotation/Proof 1
Theorem
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:
\(\ds \mathbf v\) | \(=\) | \(\ds \tuple {X, Y}\) | Definition of Vector Quantity | |||||||||||
\(\ds \tuple {X, Y}\) | \(=\) | \(\ds \tuple {x_2 - x_1, y_2 - y_1}\) | Definition of Vector Quantity |
Then we have:
\(\ds \size {\mathbf v}\) | \(=\) | \(\ds \sqrt {X^2 + Y^2}\) | Definition of Vector Length | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}\) | Substitution |
where $\size {\mathbf v}$ denotes the magnitude of $\mathbf v$.
The square of $\size {\mathbf v}$ is:
\(\ds \size {\mathbf v}^2\) | \(=\) | \(\ds \paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {x_2}^2 - 2 x_1 x_2 + {x_1}^2 + {y_2}^2 - 2 y_1 y_2 + {y_1}^2\) |
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:
\(\ds {X'}^2\) | \(=\) | \(\ds \paren {x_2 \cos \theta - y_2 \sin \theta - x_1 \cos \theta + y_1 \sin \theta}^2\) |
There are $16$ terms so we group them on four lines as a bookkeeping device:
\(\ds {X'}^2\) | \(=\) | \(\ds {x_2}^2 \cos^2 \theta + {y_2}^2 \sin^2 \theta + {x_1}^2 \cos^2 \theta + {y_1}^2 \sin^2 \theta\) | ||||||||||||
\(\ds \) | \(-\) | \(\ds 2 x_2 y_2 \sin \theta \cos \theta - 2 x_1 x_2 \cos^2 \theta + 2 x_2 y_1 \sin \theta \cos \theta\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds 2 x_1 y_2 \sin \theta \cos \theta - 2 y_1 y_2 \sin^2 \theta\) | ||||||||||||
\(\ds \) | \(-\) | \(\ds 2 x_1 y_1 \sin \theta \cos \theta\) |
Then:
\(\ds {Y'}^2\) | \(=\) | \(\ds \paren {x_2 \sin \theta + y_2 \cos \theta - x_1 \sin \theta - y_1 \cos \theta }^2\) |
Group the terms on separate lines:
\(\ds {Y'}^2\) | \(=\) | \(\ds {x_2}^2 \sin^2 \theta + {y_2}^2 \cos^2 \theta + {x_1}^2 \sin^2 \theta + {y_1}^2 \cos^2 \theta\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds 2 x_2 y_2 \sin \theta \cos \theta - 2 x_1 x_2 \sin^2 \theta - 2 x_2 y_1 \sin \theta \cos \theta\) | ||||||||||||
\(\ds \) | \(-\) | \(\ds 2 x_1 y_2 \sin \theta \cos \theta - 2 y_1 y_2 \cos^2 \theta\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds 2 x_1 y_1 \sin \theta \cos \theta\) |
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:
\(\ds \size {\mathbf v'}^2\) | \(=\) | \(\ds {x_2}^2 - 2 x_1 x_2 + {x_1}^2 + {y_2}^2 - 2 y_1 y_2 + {y_1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2\) |
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:
- $\size {\mathbf v'} = \size {\mathbf v}$
$\blacksquare$