Given three vectors in 3D space $\vec{a}$, $\vec{b}$, & $\vec{c}$ lay in the same plane.
Let $\vec{c}$ be unknown. Knowing the vectors $\vec{a}$ & $\vec{b}$ and the angle between angle between $\vec{b}$ & $\vec{c}$ how do you find $\vec{c}$?
I know from the known vectors $\vec{a}$ & $\vec{b}$ the plane the vectors lay on can be determined, additionally I know $\vec{c}$ can be any linear combination of its x,y,z components yielding the proper direction since its magnitude cannon be determined with the known information.
My question is two parts
Primary Question) How would you find $\vec{c}$?
Additional Question)if the components of each vector belong to the field of rational numbers can this be done rationally?
I have tried breaking down $\vec{a}$ into vectors parallel and perpendicular to $\vec{b}$ using vector projection and vector rejection but I was not sure how to carry on.
Best Answer
An alternative approach is, as you mention, to consider the vector $b$ and an orthogonal vector to $b$ in the same plane as $a, b$ and $c$. You can build one by considering the part of $a$ orthogonal to $b$, say \begin{equation} d = a - \frac{a \cdot b}{b \cdot b}b \, . \end{equation}
Now you can expand $c$ as a linear combination of $b$ and $d$, say \begin{equation} c = \beta \, b + \delta \, d \, . \end{equation}
Then the one additional constraint, namely the fact that $c$ is at an angle $\phi$ with $b$, can be written as \begin{equation} \frac{b \cdot c}{\sqrt{b^2 \, c^2}} = \cos \phi \,. \end{equation}
Out of the two unknowns, $\beta$ and $\delta$, we will be able to fix one of them in terms of the other (up to a sign) with this constraint. The sign arbitrariness is simply due to the fact that there are generically two distinct co-planar rays at an angle $\phi$ with $b$: one on the "right", one on the "left". The clear exceptions are when $\phi$ is an integer multiple of $\pi$, in which case there is only a single distinct ray. We can write the solution in a way that covers these cases as well, \begin{equation} \delta \, |\cos \phi| = \pm \, \lambda \, \beta \,|\sin \phi| \, , \end{equation} with \begin{equation} \lambda = \frac{b^2} {\sqrt{a^2 b^2 - \left( a \cdot b \right)^2}} \, . \end{equation}
EDIT: Let's see how to find this relation. First, remember that $d$ and $b$ are orthogonal. Hence, $b \cdot d = 0$ and then $b \cdot c = \beta \, b^2$. Orthogonality also implies that $c^2 = \beta^2 \, b^2 + \delta^2 \, d^2$. By taking this into account and squaring the constraint for $\cos \phi$, you'll find $\beta^2 \, b^2 \, \sin^2 \phi = \delta^2 \, d^2 \, \cos^2 \phi$. Then if you expand $d^2$, you'll find the relation above.