Hi Everyone,

Post your solutions to the Feb 20 In-Class Exercise to this thread.

Best,

Chris

Hi everybody,

||((1),(2),(3),(4))|| = sqrt(1^2 + 2^2 + 3^2 + 4^2) = sqrt(30) ||((-2),(1),(2),(1))|| = sqrt((-2)^2 + 1^2 + 2^2 + 1^2) = sqrt(10)

q1 \cdot q2 = ((-14), (-8), (-2), (-6))

((cos(\pi/4)), (sin(\pi/4) 1/sqrt(3) ((1), (1), (1)))) \approx ((0.71), (0.41), (0.41), (0.41))

I got √30 for the length of the first vector and √10 for the second. Their quaternionic product is (-14, -8, -2, -6).

consider the 2 quaternions: [1 2 3 4]^t [-2 1 2 1]^t

length of [1 2 3 4]^t is sqroot(30) length of [-2 1 2 1]^t is sqroot(10)

multiplication of [1 2 3 4]^t [-2 1 2 1]^t is [-14 -8 -2 -6]^t

rotation by pi/2 abt (1,1,1) vector is [cos(pi/4) (sqroot(3)^-1)sin(pi/4)[1 1 1]^t]^t

1/
||(1,2,3,4)||= sqrt(30) ||(-2,1,2,1)||= sqrt(10)

2/ Quaternionic product: (-14, -8, -2, -6)

3/ ((cos(pi/4)), ( (1/sqrt(3) sin(pi/4))(1,1,1))) = ((1/sqrt(2), ((1/sqrt(2))* (1/sqrt(3))* (1,1,1)))

1. Length of matrix 1 (m1) = 5.477 Length of matrix 2 (m2) = 3.162
2. m1*m2 = [-14 -8 -2 -6]^t <br>
3. Rotation of (1 1 1) by pi/2 = [0.707 0.577 0.577 0.577]^t

(length of quaternion [1, 2, 3, 4]) = sqrt(30) (length of quaternion [-2, 1, 2, 1]) = sqrt(10) product = [-14, -8, -2, -6] rotation by pi/2 about (1, 1, 1): [sqrt(2)/2, (sqrt(2)/2) * ((sqrt(3)/3) * (1, 1, 1))]

1. sqrt(30), sqrt(10)
2. [-14, -8, -2, -6]
3. [cos(pi/4), sin(pi/4)/sqrt(3), sin(pi/4)/sqrt(3), sin(pi/4)/sqrt(3)]