# Finding unknown 3D vector given 2 known vectors and 2 known angles

So I have a 3D vector math problem that I'm having difficulty solving. Basically I have two known vectors in the form (x,y,z), let's call them C and P, and I want to find a third unknown vector, let's call it A.

I know the angle between C and A is 60 and the angle between P and A is 90. I also know that C and P are unit vectors already and that my unknown A is going to be a unit vector. I also know that I already have two solutions to the problem (as opposed to 0 or 1 possible solutions depending). I formulated it as follows and I think this is correct. So I have:

C=(0, 0, 1)

P=(0.61, 0.61, 0.5)

A=(x, y, z)

This gives me the following three equations which should allow me to find both solutions to my three variables but I'm not getting results that make any sense.

$$\cos(90) = 0.61x + 0.61y + 0.5z$$ $$\cos(60) = 0x + 0y + 1z$$ $$\sqrt{x^2 + y^2 + z^2 } = 1$$

Finding z is easy as it must be 0.5 (which makes sense as z must be fixed for both solution in my problem). I've tried solving using substitution from here to find a quadratic equation for y, finding both y values, then using substitution again to find corresponding x values. I'm not getting any useful results however as the results aren't unit vectors and are therefore incorrect.

For clarity sake I'm trying to find the intersection points between a great circle and small circle on a 3D sphere. P represents the normal vector to the plane that forms the great circle and N represents the centre of the 60 degree small circle on the sphere.

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Vector $\vec{n} = (\vec{x} \times \vec{y})$ is perpendicular to the plan spanned by $\vec{x}$ and $\vec{y}$. Assuming that $\vec{x}$ and $\vec{y}$ are not co-linear, $\vec{n} \not= \vec{0}$.
Now $\vec{x}$, $\vec{y}$ and $\vec{n}$ can be used as a basis. Let $\vec{z} = c_1 \vec{x} + c_2 \vec{y} + c_3 \vec{n}$. It now remains to find $c_i$ from conditions that $$(\vec{z}, \vec{z}) = 1 \qquad (\vec{z}, \vec{x}) = \cos(\alpha) \qquad (\vec{z}, \vec{y}) = \cos(\beta)$$ $$(\vec{z}, \vec{z}) = c_1^2+c_2^2+c_3^2 + 2 c_1 c_2 (\vec{x}, \vec{y})$$ $$\cos(\alpha) = (\vec{z}, \vec{x}) = c_1 + c_2 (\vec{x}, \vec{y}) \qquad \cos(\beta) = (\vec{z}, \vec{y}) = c_2 + c_1 (\vec{x}, \vec{y})$$ From here: $$c_1 = \frac{\cos(\alpha) - \cos(\beta) (\vec{x}, \vec{y})}{1-(\vec{x}, \vec{y})^2} \quad c_2 = \frac{\cos(\beta) - \cos(\alpha) (\vec{x}, \vec{y})}{1-(\vec{x}, \vec{y})^2}$$ $c_3^2$ is then easily determined: $$c_3^2 = 1 - \frac{\cos^2(\alpha) + \cos^2(\beta) - 2 \cos(\alpha) \cos(\beta) (\vec{x}, \vec{y})}{1-(\vec{x}, \vec{y})^2}$$ The sign of $c_3$ is not determined by your requirements.