I am given a 3-component vector $\vec v$. There are obviously an infinite number of orthogonal vectors to $\vec v$. I need to find the specific orthogonal vector, lets call it $\vec{x}$, in the plane defined by $\vec v$ and the positive z-unit vector.
Is there some sort of standardized equation to find $\vec x$?
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If $\vec{v}=(v_1,v_2,v_3)^T$ all the vectors $\vec{x}=(x,y,z)^T$ orthogonal to $v$ are on the plane: $$\vec x \cdot \vec v=(x,y,z)\cdot(v_1,v_2,v_3)^T=0$$ so there are vectors orthogonal to $\vec v$ vith any component $x$ , $y$ or $z$ as great as we want.
The plane containing $\vec k = (0,0,1)^T$ and $\vec v$ has equation: $\vec x \cdot(\vec k \times \vec v)=0$ i.e. $v_2x-v_1y=0$. intersect this plane with the plane orthogonal to $\vec v$ (i.e. $ v_1x+v_2y+v_3z=0$) and you find all vectors orthogonals to $\vec v$ and in the plane containing $\vec k$ and $\vec v$.
Without loss of generality (i.e., up to overall scaling), you're looking for a vector of the form $$ x = cv + e_{3} = cv + (0, 0, 1) $$ orthogonal to $v$ for some (unknown) scalar $c$. Taking the dot product with $v$ gives $0 = c(v \cdot v) + v \cdot e_{3}$, or $$ c = -\frac{v \cdot e_{3}}{v \cdot v}. $$ (This presumes $v \neq 0$, but if $v = 0$ the vector $x = e_{3}$ suits your needs.)