Question

Previously you found that a 15-degree counterclockwise rotationcenteredat(2,1)sendsthepoint(4,6)toanother point (x, y) ⇡ (2.638, 6.347). The diagram on the right shows the vector v in the same direction as vector [2,5], only with a di↵erent length. Similarly, the vector u is per- pendicular to v and a di↵erent length. Show how you can use (2,1)+v+u to find the rotated point (x,y).

Answer 1

sorry it's formatted weirdly

Answer 2

Yesit'scertainlyweird,butthequestionisquiteinteresting!

Let
C=(2,1) [centre of rotation]
P=(4,6) [point to be rotated]
t=π/12 [15°, angle of rotation]

The first part of the problem is solved as follows:
Rotated point, P', is obtained by
1. translating the point P so that C becomes the origin,
P->(P-C)
2. roatate by π/12 clockwise
multiply by the rotation matrix, R(t):
| cos(t) -sin(t) |
| sin(t) cos(t) |
P->R(t)P
3. reverse the translation
P->P+C

Summing up, the rotation procedure is
P'=C+R(t)(P-C)
=(2cos(π/12)-5sin(π/12)+2, 5cos(%pi;/12)+2sin(π/12)+1)

Second part of the question.
It turns out that the vector <2,5> is exactly the vector
P-C=<4-2,6-1>=<2,5>
and it turns out that the two rows of the rotation matrix
<cos(t), -sin(t)> and
<sin(t), cos(t)>
are orthogonal (perpendicular).

This means that
R(t)<P-C>=
|2 5| <cos(t),-sin(t)>
|5 -2|
=cos(t)<2,5>-sin(t)<5,-2>
=u+v
By translating back, we get
(2,1)+u+v