Intuition - dot product doesn't change under rotation

Question

For definiteness let $a:(x_1,y_1),b:(x_2,y_2)$ be unit position vectors:
$$a \cdot b = x_1x_2+y_1y_1 = \cos (\phi)$$

where $\phi$ is the angle between $a,b$.
Clearly the angle $\phi$ doesn't change when we rotate the plane.
However I'm finding it hard to see how the sum of products $x_1x_2+y_1y_2$ remains constant.

I feel an explanation using coordinate geometry helps; that is by manipulating $x,y$ coordinates of position vectors $a,b$, how to see the following product remains constant? $$\cos a\cos b +\sin a \sin b$$

Answer 1

If $(x_1,y_1)\mapsto(\cos(\vartheta)x_1-\sin(\vartheta)y_1,\sin(\vartheta)x_1+\cos(\vartheta)y_1)$ under an anticlockwise $2$-d rotation matrix by angle $\vartheta$, and likewise for $(x_2,y_2)$, then, calling the new points $a',b'$:

$$\begin{align}a'\cdot b'&=[(\cos(\vartheta)x_1-\sin(\vartheta)y_1)\cdot(\cos(\vartheta)x_2-\sin(\vartheta)y_2)]\\&+[(\cos(\vartheta)y_1+\sin(\vartheta)x_1)\cdot(\cos(\vartheta)y_2+\sin(\vartheta)x_2)]\\&=[\cos^2(\vartheta)x_1x_2+\sin^2(\vartheta)y_1y_2-\sin(\vartheta)\cos(\vartheta)(x_1y_2-x_2y_1)]\\&+[\cos^2(\vartheta)y_1y_2+\sin^2(\vartheta)x_1x_2+\sin(\vartheta)\cos(\vartheta)(x_1y_2+x_2y_1)]\\&=(\cos^2(\vartheta)+\sin^2(\vartheta))x_1x_2+(\cos^2(\vartheta)+\sin^2(\vartheta))y_1y_2\\&=x_1x_2+y_1y_2\\&=a\cdot b\end{align}$$

If the very first step is confusing, study carefully the rotation described below:

$$\begin{pmatrix}\cos(\vartheta)&-\sin(\vartheta)\\\sin(\vartheta)&\cos(\vartheta)\end{pmatrix}\begin{pmatrix}x_1\\y_1\end{pmatrix}$$

Answer 2

Since $a$ and $b$ are unit vectors, we can find $\alpha$ and $\beta$ such that $a = (\cos\alpha, \sin\alpha)$ and $b = (\cos\beta, \sin\beta)$. Then $$ a \cdot b= \cos\alpha\,\cos\beta + \sin\alpha\,\sin\beta = \cos(\beta-\alpha) $$ So, the dot product depends only on the angle $\beta-\alpha$ between the two vectors.