# Question concerning cross product and orthonormal vectors

Assume we have a vector $u= (u_1.u_2, u_3) \in R^3$

My problem is to find vectors $\vec w, \vec v$ such that $u= v \times w$ All vectors should be orthonormal.

If $u= (u_1, u_2, u_3)$ ,is there a way to express these vectors $\vec w, \vec v$ with respect to $\vec u$.

If there is, I would like to see an example.

I am supposed to build a matrix whose columns consist of $A =(\vec v, \vec w, \vec u)$ Remember that all vectors should be orthonormal to each other, and that $u= v \times w$

This matrix A will represent a rotation matrix, an element $\in SO(3)$

My goal is to show that $R= ASA^T$ (S a general rotation matrix around z-axis) is again a an element $\in SO(3)$, a rotation around the unit axis $u$, on the the form as you see in this article:

http://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle

• I updated my answer below, but you can actually show your goal without looking at matrix columns or entries, and just using the characterization of matrices $B \in SO(3)$ as those for which ${}^t B B = 1$ and $\det B > 0$. Commented Oct 20, 2014 at 22:01

Pick any vector $v$ orthonormal to $u$; one way to do this is to pick any vector $v'$ not parallel to $u$ and then apply the Gram-Schmidt algorithm to the pair $(u, v')$. Then, set $w := u \times v$, so that $(u, v, w)$ is an oriented orthonormal basis of $\mathbb{R}^3$; in particular, $(v, w, u)$ is also an oriented orthonormal basis of $\mathbb{R}^3$ and so $$u = v \times w$$ as desired. (If one isn't familiar with this characterization of such bases, we can alternately see this with the iterated cross product identity $$a \times (b \times c) = b \cdot (a \times c) - c \cdot (a \times b).\text{)}$$

One can show your desired result (that $R \in SO(3)$) quickly using the characterization of matrices $B \in SO(3)$ as exactly those for which ${}^t B B = I$ and $\det B > 0$.

• But OP also wants orthonormality (not just orthogonality). Commented Oct 20, 2014 at 12:32
• Ah, I didn't see that before---that makes the argument easier, and I've updated the answer. Commented Oct 20, 2014 at 22:00

If $u_1=0$, let $v=(1,0,0)$ and let $w=u\times v$.

If $u_2=0$, let $v=(0,1,0)$, etc.

Otherwise, let $v=(u_2,-u_1,0)/\sqrt{u_1^2+u_2^2}$, etc.

• So if we have that $u=(u_1, u_2, u_2)$ and $v = u_2, -u_1, 0)$, how to we choose w? Commented Oct 21, 2014 at 7:01
• @Eirik If those $u, v$ are orthonormal then $u_2 = 0$, which forces $u = \pm 1$. If you meant $u = (u_1, u_2, u_3)$, this forces $u_3 = 0$. In either case, we get $w = u \times v = (0, 0 , -1)$. Commented Oct 21, 2014 at 7:28
• "Etc." means "Let $w=u\times v$." So long as $u$ and $v$ are orthogonal, their cross product will be orthogonal to both of them. And so long as $u$ and $v$ are in addition unit vectors, their cross product will be a unit vector. Commented Oct 21, 2014 at 9:32
• @Travis, I meant $u=(u_1, u_2, u_3)$. But if $u_3=0$ I dont see how you can obtain the rotation matrix in the link I posted. Here, $R$ consists of all the coordinates $u_1, u_2, u_3$ which corresponds to $u_x, u_y, u_z$. This is the matrix that I am supposed to get to multiply the matrices $ASA^T$ as mentioned above. Then The matrix A has to contain all the parameters $u_1, u_2, u_3$, since the matrix S just contains cosines and sines. Commented Oct 21, 2014 at 11:05
• I want a matrix $A= (\vec v, \vec w, \vec u)$ which satisfies the that the vectors are orthonormal and $u=v \times w$ and $ASA^T$ is equal to the matrix in the link I posted. If you just can give tell how this matrix A looks like. Commented Oct 21, 2014 at 11:10