Vector Magnitude is Invariant Under Rotation

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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 1

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:

$\size {\mathbf v'} = \size {\mathbf v}$

$\blacksquare$


Proof 2

We offer three equivalent statements:

By definition, rotation of the coordinate system affects the coordinates and not the vector.


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Rotation is a rigid transformation. It does not change side lengths or angles.

The equations of rotation of coordinates are linear transformations.

$\blacksquare$

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