How to rotate vector A around another vector B?

Question

Lets say I know what is directly above a turret. I also know where the turrets gun is currently pointing. I would like to know where will the turret's gun will point if I rotate it.

I suppose I could multiply \$A \cdot B\$, project \$A\$ on the plane \$B\$ is a normal of and then do the rotation in 2D and add back \$A \cdot B\$ but I am sure the math should be simpler than that?

Answer 1

You want to reflect \$\vec A\$ across \$\vec B\$. This has a simple formula, if \$\vec B\$ has length 1:

$$-\vec A + 2( \vec A \cdot \vec B)\vec B$$

What we've done here is to reverse \$\vec A\$ first, which means we negated both the component of \$\vec A\$ that's parallel to \$\vec B\$ and the component of \$\vec A\$ perpendicular to \$\vec B\$. Then we added twice the parallel component, \$-1 + 2 = +1\$, bringing it back to its previous value. So the net effect is that the part of \$\vec A\$ that was parallel to \$\vec B\$ remains unchanged, and only the perpendicular part gets reflected.

Answer 2

You simply want to apply the rotation matrix.

Assuming you use a cartesian metric for your game, and your gun is pointing in direction \$ \vec{A} = (x,y,z) \$, and you rotate around the vertical axis \$ \vec{B} = (0, 0, 1) \$. Then the rotation \$ R_z \$ by and angle \$ \theta \$ is defined as $$ R_z(\theta) = \left(\begin{array}{ccc}\cos \theta, & -\sin \theta,& 0\\\sin\theta, &\cos \theta,& 0\\0,&0,&0 \end{array}\right) $$

Thus the rotated vector \$\vec{A}^\prime = \vec{A} \cdot R_z = (x\cos\theta-y\sin\theta, x\sin\theta + y\cos\theta, 0)\$

For a more elaborate treatise around general axis, consult a book on algebra or have a look on the wikipedia article on rotation:

The matrix of a proper rotation \$R\$ by angle \$\theta\$ around the axis \$\vec u =(u_{x},u_{y},u_{z})\$, a unit vector with \$u_{x}^{2}+u_{y}^{2}+u_{z}^{2}=1\$, is given by:[3]

$$\small {R = \begin{bmatrix}\cos \theta +u_{x}^{2}\left(1-\cos \theta \right)&u_{x}u_{y}\left(1-\cos \theta \right)-u_{z}\sin \theta &u_{x}u_{z}\left(1-\cos \theta \right)+u_{y}\sin \theta \\u_{y}u_{x}\left(1-\cos \theta \right)+u_{z}\sin \theta &\cos \theta +u_{y}^{2}\left(1-\cos \theta \right)&u_{y}u_{z}\left(1-\cos \theta \right)-u_{x}\sin \theta \\u_{z}u_{x}\left(1-\cos \theta \right)-u_{y}\sin \theta &u_{z}u_{y}\left(1-\cos \theta \right)+u_{x}\sin \theta &\cos \theta +u_{z}^{2}\left(1-\cos \theta \right)\end{bmatrix}}$$