matrix representation of a trigonometric rotation

Question

Hey guys!I have a couple of doubts regarding this exercise, for a) I think that the Matrix

rotation of P is [(cos t, -sen t) , (-sen t, cos t)] and for Q [(-cos t, -sen t), ( sen t,

cos t)] , is that right?

Regarding B... no idea..and I know c and d.

Thanks very much in advance :)

Answer 1

For a.) $P$ is the image of $i$ so it should have two coordinates: $P=(\cos t,\sin t)$. Also $Q=(-\sin t,\cos t)$. The matrix of the rotation $R$ is $$\begin{pmatrix}\cos t & -\sin t \\ \sin t& \cos t\end{pmatrix}.$$ Now, b.) is clear.

Answer 2

Well, in the picture, $P$ is a vector, namely $\pmatrix{\cos t\\ \sin t}$, and also is $Q$, namely $\pmatrix{-\sin t\\ \cos t}$ -- recheck the signs, the picture helps.

Then $R(t)$ is the matrix with columns $\pmatrix{P & Q}$. This should also clarify b).

Answer 3

For a) there is only one matrix for both $P$ and $Q$. $P$ is the rotation of $i$ and $Q$ is the rotation of $j$. You should only have a minus sign on one of the $\sin$ entries.

For b) you can just do the arithmetic indicated and see that it gives the matrix $R(t)$