[1 1 1] is the normal vector that describes the plane in R3 If the plane is rota

Question

[1 1 1] is the normal vector that describes the plane in R3

If the plane is rotated by , find the standard matrix that gives this transformation.

In other words... find A such that-

T([x y z])=A[x y z]=new vector rotated about plane by .

T is the tansformation function

A is the standard matrix

Explanation / Answer

Derivation of 3d rotation matrix: rotation of vector r theta degrees and around vector n and becoming vector s rp = r-parallel = n * r = sp = s-parrallel = n*s (since rotation doesn’t change length) rPerp = r-Perpendicular and sPerp = s-Perpendicular Find s: Since I know r, the goal is to get everything in terms of r. s = sp + sPerp => s = rp + sPerp To find sPerp, let’s create a plane that sPerp is on with two basis that I know. I can get one (call it V) by doing a cross product (?) with n and rPerp, which will produce a vector that is perpendicular to rPerp and n. So our two new basis can be rPerp and this new vector V. ?=theta with some good old trig I get: sPerp = |sPerp|*cos(?)*unit_rPerp + |sPerp|*sin(?)*unit_v Since sPerp and rPerp are on a circle, I know that their lengths are the same. Also I know the length of V is the same as rPerp because of a property of the cross product. Namely, |n ? rPerp| = |n|*|rPerp|*sin(alpha), n’s length is one and the angle betIen them is 90 degrees and sin(90)=1, thus |V|=|rPerp| This lets us write sSperp like this: sPerp = |rPerp|*cos(?)*unit_rPerp + |v|*sin(?)*unit_v sPerp = cos(?)*rPerp + sin(?)*V Rewrite s: s = rp + cos(?)*rPerp + sin(?)*V before I write everything in terms of r, note that (n ? rPerp) is equal to (n ?(r-rp)) => (n?r - n?rp) and since rp is in the same direction as n, n?rp is the 0 vector, and I’re left with n?r. so V is also n?r! s = (n? r)* n + cos(?)*(r - (n? r)* n) + sin(?)*( n?r) Using the tensor product I can rewrite (n? r)* n as M*r, where M is matrix that when multiplied with r produces (n? r)* n. r = (rx,ry,rz) and n= (nx,ny,nz) (n? r)*n=((rx*nx+ry*ny+rz*nz)*nx, (rx*nx+ry*ny+rz*nz)*ny, (rx*nx+ry*ny+rz*nz)*nz) (rx*nx*nx+ry*ny*nx+rz*nz*nx, rx*nx*ny+ry*ny*ny +rz*nz*ny, rx*nx*nz+ry*ny*nz +rz*nz*nz) => = M*r so now I have: s = M*r + cos(?)*(r - M*r) + sin(?)*( n?r) s = M*r + cos(?)*r – cos(?)*M*r + sin(?)*( n?r) s = (1-cos(?))*M*r + cos(?)*r + sin(?)*( n?r) Next I’ll write n?r also as a matrix (say P) multiplied by r. n?r = =>=P*r So now I have: s = (1-cos(?))*M*r + cos(?)*r + sin(?)* P*r since (1-cos), cos and sin are scalars, simply make them scalar matrices! s= [(1-cos(?))*M + cos(?)*I + sin(?)* P]*r Rotation Matrix = [(1-cos(?))*M + cos(?)*I + sin(?)* P] Which is: => thus I have the Rotation matrix! Note: If I plug in the basis vectors from the 3D Cartesian coordinate system I’ll get the three rotation matrices that rotate about an axis. Put n = {0,0,1} in and we get: => => which is the Rotation matrix about the z axis! Put n = {0,1,0} in and we get: => => which is the Rotation matrix about the y axis! Put n = {1,0,0} in and we get: => => which is the Rotation matrix about the x axis! Note: That if you rotate -? degrees about –n you will have the same rotation as rotating ? degrees about n. Put your eye at Q and rotate +? degrees is counterclockwise and -? degrees is clockwise, and you will be doing the same rotation. Proposition to Get Rid of Trig and Square Root calls If I know vector s (the final vector from the rotation) then I can easily get rid of the trig and square roots. Let vector V = r?s and remember that vector n=r?s/(|r?s|) Cos(?)=r?s/(|r|*|s|), |r|=1, |s|=1 => Cos(?)=r?s Sin(?)=|r?s|/(r||*|s|), |r|=1, |s|=1 => Sin(?)=|r?s| |V|=|r?s|==Sin(?) V?V= n.x = , n.y = , n.z = n.x*n.x = => , n.y*n.y = , n.z*n.z = , n.x*n.y = => , n.x*n.z = , n.y*n.z = n.z*Sin(?) = *|r?s| => *|V| => V.z n.y*Sin(?) => V.y n.x*Sin(?) => V.x Now I can rewrite the rotation matrix in terms of vector V, which will be without trig and square roots.

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