Question

Problem: Given a rotation R of R3 about an arbitrary axis through a given angle find the matrix which represents R with respect to standard coordinates. Here are the details: The axis of rotation is the line L, spanned and oriented by the vector v (1,一1,-1) . Now rotate R3 about L through the angle t = 4 π according to the Right 3 Hand Rule Solution strategy: If we choose a right handed ordered ONB B- (a, b,r) for 3 R", with r the unit vector in the direction of v, then L appears like the z-axis in standard coordinates. Therefore cost -sint 0 sint cost 0 B-B 0 0 Now, all we need to do is convert to standard coordinates. This is particularly, easy because the matrix Cse a br which converts from standard coordinates to B-coordinates is an orthogonal matrix. So, Si-B Si-B Therefore RI=CS, BİRIB..BC Solution: For some of the calculations, you may wish to use a calculator. Enter root expressions, such as 5 as "sqrt(5)". Do not replace a root expression by an approximate ratio or and approximate decimal number. - The system will mark those approximations "wrong". Step 1: Construction of B. The unit vector in the direction of v is A unit vector perpendicular to r is: b = (1, 1,0). Therefore, a vector perpendicular to both b and r is a b x r; i.e. a= Indeed, llal| 1, as this is the area of the square spanned by the unit vectors b and r. Further, the resulting ONB B-(a, b, r) represents represents the right hand orientation of IR because Thus we find: and and, therefore Check

Answer 1

Unit vector in the direction of v, lv -V12 (-1)2 + (-1)2 키이 = V3 V3 0

0 уб уб

アで,7 흖흖 6 0 6 アアーア V6 カー V3

「cos停) -sin(,) 01 4 π COS 0 0

$\hspace{-0.6cm}[R]_{B\leftarrow B} =\begin{bmatrix} -\frac{1}{2} &\frac{\sqrt3}{2} & 0\\ &&\\ -\frac{\sqrt3}{2} & -\frac{1}{2}& 0\\ &&\\ 0 & 0& 1 \end{bmatrix}\\\\$

$\hspace{-0.6cm}[R] = C_{S\leftarrow B}[R]_{B\leftarrow B}C_{S\leftarrow B}^T\\$

0 0 0 2 0

010 -00 00-1