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madeleinejames20 · Accepted Answer

a) To find the matrix that represents the rotation in 3D with a rotation around the z-axis by an angle of π4 followed by a rotation around the y-axis by an angle of π6, we can multiply the individual rotation matrices.The rotation matrix around the z-axis by an angle θ is given by:Rz(θ)=[cos(θ)−sin(θ)0sin(θ)cos(θ)0001]The rotation matrix around the y-axis by an angle ϕ is given by:Ry(ϕ)=[cos(ϕ)0sin(ϕ)010−sin(ϕ)0cos(ϕ)]Substituting the given angles, we have:Rz(π4)=[cos(π4)−sin(π4)0sin(π4)cos(π4)0001]=[22−22022220001]Ry(π6)=[cos(π6)0sin(π6)010−sin(π6)0cos(π6)]=[32012010−12032]Now, we multiply the matrices to get the final rotation matrix:R=Ry(π6)·Rz(π4)R=[32012010−12032]·[22−22022220001]Evaluating the matrix product, we obtain:R=[32·22−32·221222220−32·2232·2232]=[64−641222220−646432]Therefore, the matrix that represents the rotation in 3D, consisting of rotating around the z-axis by π4 followed by rotating around the y-axis by π6, is:R=[64−641222220−646432]b) To find the matrix that represents the rotation in 3D, consisting of the same rotations as in part (a), but in the other order, we multiply the individual rotation matrices in reverse order:R′=Rz(π4)·Ry(π6)Substituting the given angles and evaluating the matrix product, we obtain:R′=[22−22022220001]·[32012010−12032]Evaluating the matrix product, we obtain:R′=[22·32−22·1222·1222·022·122·022·−1222·022·32]=[64−24240220−24034]Therefore, the matrix that represents the rotation in 3D, consisting of rotating around the y-axis by π6 followed by rotating around the z-axis by π4, is:R′=[64−24240220−24034]c) To act by the matrix of part (a) on the unit vector i in the positive x-direction, we multiply the matrix R with the column vector [100].R·[100]=[64−641222220−646432]·[100]Evaluating the matrix-vector product, we obtain:R·[100]=[6422−64]We can check that the resulting vector is still a unit vector by calculating its magnitude:‖[6422−64]‖=(64)2+(22)2+(−64)2=616+24+616=34=32The magnitude of the resulting vector is 32, which is equal to 1. Therefore, the resulting vector is still a unit vector.Therefore, when we act by the matrix of part (a) on the unit vector i in the positive x-direction, the resulting vector is [6422−64], which is still a unit vector.

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