Rotation of 4D vector

Question

In an XYZU coordinate system, there are 6 planes of rotation. I have the vector: $$[1,1,1,1]$$ I want to rotate it to be aligned with $$[0,0,0,1]$$ by applying rotations in XY, XZ, YZ, XU, YU, and ZU planes.

I believe it only takes rotations in the XU, YU, and ZU planes to achieve this.

Although, I am having trouble figuring out how to calculate these angles.

Any help would be appreciated.

My Attempt:

In 3D, If I wanted to rotate [1,1,1] to [0,0,1] It would take a rotation of $\frac{\pi}{4}$ in the XZ plane and a rotation of $arcsin(1/\sqrt{3})$ in the YZ plane. I was able to figure this out because I can picture things in 3D! But I'm having trouble generalizing this to 4D. Since the magnitude of [1,1,1,1] is 2, I know that I will need to include "2" instead of a $\sqrt{3}$, but not sure how to perform the vector algebra to generalize to 4D.

Answer 1

Note that the angle between any two vectors satisfies $$ \theta_{v,w} = \arccos \left(\frac{\langle v,w \rangle }{\|v\| \,\|w\|}\right) $$ where $\langle v,w \rangle$ denotes the "dot-product" of $v$ and $w$.

With that in mind, we can bring $(1,1,1,1)$ to the desired orientation with three rotations as follows:

1. Take $(1,1,1,1) \to (0,\sqrt{2},1,1)$ with a $\pi/4$ rotation in the $XY$-plane
2. Take $(0,\sqrt{2},1,1) \to (0,0,\sqrt{3},1)$. The angle of this rotation in the $YZ$ plane is $\arccos(\frac{\sqrt{3} + 1}{4})$.
3. Take $(0,0,0,\sqrt{3},1) \to (0,0,0,2)$. The angle of this rotation in the $ZU$ plane is $\arccos(\frac{2}{4}) = \pi/6$.