Finding a vector in 3 dimensions

Question

How could you find a vector if you know the values of another vector and the angle between them in three dimensions? Also, the y component is the same for both vectors and the magnitude of both vectors is equal to 1.

Answer 1

${\bf a} = (a_x, a_y, a_z)$ (all known)

${\bf b} = (b_x, a_y, b_z)$ (not known)

Normalization:

$| {\bf a}| = 1 \to {\bf a} = (a_x, a_y, \sqrt{1-a_x^2 - a_y^2})$ (all known)

$|{\bf b}| = 1 \to {\bf b} = (b_x, a_y, \sqrt{1 - b_x^2 - a_y^2})$ (not all know)

$\cos \theta = {\bf a} \cdot {\bf b} = a_x b_x + a_y^2 + \sqrt{1 - a_x^2 - a_y^2} \sqrt{1 - b_x^2 - a_y^2}$ (known)

One equation in one unknown variable, $b_x$.

Answer 2

You have your known vector. We denote this $\vec a=(a_1,a_2,a_3)$. We want to find the vector $\vec b=(b_1,b_2,b_3)$, where we know $a_2=b_2$, and the angle between them, $\theta$, is known. We also know $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}=1$ and $|\vec b|=\sqrt{b_1^2+b_2^2+b_3^2}=1$. We can utilise these facts when we compute the dot product.$$\begin{align}\vec a\cdot \vec b&=|\vec a||\vec b|\cos\theta=\cos\theta\\&=a_1b_1+a_2^2+a_3b_3\end{align}$$

This gives us two equations for $b_1,b_3$: $$b_1^2+b_3^2=1-a_2^2\\a_1b_1+a_3b_3=\cos\theta-a_2^2$$ These equations can then be solved (e.g. by substitution).