I am looking for some performant algorithm for a rotation of multiple points in $\mathbb{R}^2$ (and maybe also $\mathbb{R}^3$ with one coordinate zero) around the origin.

Full story: I want to draw a texture on a quad plane along some line, rotated to the direction of the line. I know how to perform the translation beside the rotation.

If you already know the angle, you can build a rotation matrix or a quaternion/complex number using that for each point.

However, what I need is just a transformation between two direction vectors in the special case where the source direction is up $\vec{s}=\pmatrix{0\\1}$. The destination direction can be arbitrary and the vector does not necessarily have to be normalized ($\lvert\vec{d}\rvert \not=1 $).

Can it be done by getting the normalized unit vector $\vec{u_d} = \frac{\vec{d}}{\lvert\vec{d}\rvert} = \frac{\pmatrix{x\\y}}{\sqrt{x^2+y^2}}$ and just using the coordinates as $\sin$ / $\cos$ in a rotation matrix?

Since the source direction is up instead of along the x-axis, I guess $-\sin = x$ ; $cos = y$ of the normalized destination vector.

Would the thoughts above work? Is there any more performant way to do a rotation directly avoiding both, square root and sin/cos (e.g. using complex numbers or quaternions)? How else would I achieve what I want?

  • $\begingroup$ I'm neither a math pro nor a native English speaker. Math language differs between countries. Please feel free to edit my question to transform it into a fluent English math language. $\endgroup$ Jun 14, 2022 at 0:08
  • 1
    $\begingroup$ Your way seems about as simple and efficient as you can get. You might need to try swapping $x$ and $-x$ to get it to work correctly. $\endgroup$
    – David K
    Jun 15, 2022 at 12:55

1 Answer 1


As you proposed, you can use a rotation matrix $R_\theta=\left[\matrix{\cos(\theta)& -\sin(\theta)\\ \sin(\theta)& cos(\theta)}\right]$.

If you have uv coordinates $\vec{s}_1$ and $\vec{s}_2$ in the ordinary base $B=\left[\matrix{1 & 0\\ 0 & 1}\right]$ to rotate them you need to multiply them by $R_\theta$. If they concide with the base, it will result in $R_\theta$: $$R_\theta B=R_\theta,$$ meaning your rotated ${\vec{s}_1}^\theta$ and ${\vec{s}_2}^\theta$ will be $${\vec{s}_1}^\theta=\left[\matrix{\cos(\theta)\\ \sin(\theta)}\right]$$ and $${\vec{s}_2}^\theta=\left[\matrix{-\sin(\theta)\\ \cos(\theta)}\right].$$


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