# Ax = b when b = [0; 0; 0; 0]

1. Jul 30, 2008

### Four

1. The problem statement, all variables and given/known data
given z = (0, 0, 0, 1), and v = (0, x, y, z),

and the following properties hold

$$v \times u = L$$
$$L \times v = u$$
$$u \times L = v$$

v, u, l, and z are unit quaternions

In other words; we define forward (v), up (u) and right(L), find a unit quaternion that when you apply it on z, you get v, in the proper orientation.

solve for q = (a, b, c, d)

2. Relevant equations
(equation to solve for q)
$$qzq^{-1} = v$$
Left side applys q on z and right side is v

3. The attempt at a solution
$$qz = vq$$
$$dk = (xi + yj + zk)(a + bi + cj + dk)$$

after expanding and grouping the terms on the same axis together. In a matrix form I got

$$\left( \begin{array}{cccc} x & y & z & 0 \\ 0 & z & y & x \\ z & 0 & -x & y \\ -y & x & -1 & z \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \\ d \end{array} \right) = \left( \begin{array} {c} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$$

[a, b, c, d] = [0, 0, 0, 0] ? That doesn't seem right?

Shouldn't I need to use atleast the vector u?