Suppose $(x,y) \in \Bbb R^2$. Suppose we rotate this point about a vector $(a,b) \in \Bbb R^2$ through an angle $\theta$. Find the coordinate of the new point.
What I have done is as follows $:$
I first translate the vector $(a,b)$ suitably so that it becomes the origin in the new coordinate system. Let $(x',y')$ be the new coordinate of $(x,y)$. Then $x'=x-a$ and $y'=y-b$. Now in order to find out the coordinate of the point $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $\theta$ we need to rotate $(x',y')$ about the origin in the new coordinate system through the same angle $\theta$. After rotating let the coordinate of $(x',y')$ be changed to $(x'',y'')$ in the new coordinate system. Then $x'' = x' \cos \theta - y' \sin \theta$ and $y'' = x' \sin \theta + y' \cos \theta$. Replacing $x'$ and $y'$ by $x-a$ and $y-b$ respectively we get the coordinate of $(x'',y'')$ in the old coordinate system which is the required coordinate of $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $\theta$. So
$$x'' = (x-a) \cos \theta -(y-b) \sin \theta.$$
$$y'' = (x-a) \sin \theta + (y-b) \cos \theta.$$
is the required coordinate of the new point. But the answer given in my book is not matching with the above one. It is given as
$$x'' = (x-a) \cos \theta - (y-b) \sin \theta + a.$$ $$y'' = (x-a) \sin \theta + (y-b) \cos \theta + b.$$
Where have I done mistake? Please help me in this regard.
Thank you very much.
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-a\implies p=x''+a;y''=q-b\implies q=y''+b$.