2.4.4 Rotating a vector, revisited

2.4.4 Rotating a vector, revisited

[maggiemyers]

Notice that the video is part of a series of videos, notes, and problem sets. The purpose is to revisit rotation as an example of linear transformation and link it to its matrix representation. For more details, look at the free to audit materials for LAFF on edX.

[oj]

Clearest explanation of this subject thank you

[Chrysaries]

Excellent, straight-to-the-point video!

[andrew_ng]

Got my mind sorted here. Just great material

[farhanfouadacca]

Much better than 7 pages of my crap text book. Good job!

[OKJazzBro]

If anyone found it hard to understand why the rotations on the basis vectors can be used to construct the rotation on the original vector, it's because they proved using geometry in the first week's lecture that rotation is a linear transformation.
So R(v) = R(mag_x * [1, 0] + mag_y * [0, 1]) = R(mag_x * [1, 0]) + R(mag_y * [0, 1]) = mag_x * R([1, 0]) + mag_y * R([0, 1]).
Now, again using geometry they derive what R(basis vector) is for both basis vectors, which turns out to R([0, 1]) = [cos(theta), sin(theta)] and R([1, 0]) = [-sin(theta), cos(theta)]. Now if we put these back in the above formula, we'll get the full formula for the rotation of any 2D vector v.

[rabbitcreative]

"We know that", "we know that", "we know that". Speak for yourself.

[rayo]

thats mind blowing.... linear algebra is its own universe

[LAFFutX]

Hi Gary. The rotation take the vector into the second quadrant, so exactly one of the coordinates must be negative. The magnitude is reasoned to be sin( theta ). But the coordinate must be negative because of where it ends up. Hopefully this helps.



We just made a similar video for our Advanced LAFF (graduate level) course:






(Project in progress)

[leonmckoy]

needs more explanation
this did helped the geeks, not the want to learn folks.

[kasandrop]

and i still dont understand. Example could be helpful.