# Given a start point in 3d and a quaternion and length to Point B can you find Point B

Let's assume I have a start point A (x, y, z). Now the object has moved and the new orientation is given by a quaternion Q and it's pointing at point B which is L length away from it. How can I calculate the coordinates of point B(x, y, z).

You can assume that it doesn't move at all in a linear matter it just rotates. From my understanding you can get a direction from the quaternion. Now I am assuming that this a straight forward calculation. If it isn't can you point me in which direction I should look.

Apologizes in advances I am not a mathematician.

I'm not sure to well understand your question.

I start from the fact that, if we have two points $A$ and $B$, there are infinitely many rotations $R$ in $\mathbb{R}^3$ such that $R(A)=B$. In fact, for any point $P$ such that $\overline{PA}=\overline{PB}$, we can use as axis of the rotation the stright line $r$ from $P$, orthogonal to the plane that contains $A$,$B$ and $P$, and as angle of rotation the angle $\angle APB$.

So the problem is well defined only if we choose some point $P$ and, without loss of generality, we can assume that this point is the origin $O$ and $\overline{OA}=\overline{OB}$.

Now consider the vectors $\vec{a}=\vec{OA}=(x_A,y_A,z_A)^T$ and $\vec{b}=\vec{OB}=(x_B,y_B,z_B)^T$. The rotation that gives $\vec b$ from $\vec a$ is the rotation that has as axis the straight line that contain the vector $\vec w= \vec a \times \vec b$ (the cross product of the two vectors that is orthogonal the the plane containing these) and of angle $\theta=\arccos \left(\dfrac{\vec a \cdot \vec b}{|\vec a||\vec b|}\right)$ (with some care in the chose of $\theta$).

This rotation can be represented by means of quaternion. Let $$\vec u=(u_1,u_2,u_3)^T= \frac{\vec w}{|\vec w|}$$ the versor of the axis of rotation, than the rotation is given by: $$\vec b = R_{\vec u, \theta}(\vec a)=e^{\frac{\theta}{2} \vec u}\,\vec a\, e^{-\frac{\theta}{2} \vec u}$$ where $$e^{\frac{\theta}{2} \vec u}=\cos \left(\frac{\theta}{2} \right)+\vec u \sin\left(\frac{\theta}{2} \right)=\cos \left(\frac{\theta}{2} \right)+\left( u_1\mathbf i +u_2 \mathbf j +u_3 \mathbf k \right)\sin\left(\frac{\theta}{2} \right)$$ is a unitary quaternion. Note that the rotation is the same if we orient the versor in the opposite direction and we take the opposite of $\theta$.

Conversely, if we have a quaternion of the form $e^{\alpha \vec v}$ for some versor $\vec v$, than the product: $$e^{\alpha \vec v}\,\vec a\, e^{-\alpha \vec v}$$ is the rotation of $\vec a$ around the axis $\vec v$ by an angle $2\alpha$.

I hope this can be helpful, and you can see my answer to Representing rotations using quaternions for other references.