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I am trying to find all unit vectors orthogonal to $\vec{v}=\langle{3,4,0}\rangle$.
So, in this pursuit, I considered the possibility of a property of vectors:
For a given non-zero vector $\vec{v}=\langle{a,b,c}\rangle$ there exists an orthogonal vector $\vec{u}=-\langle{\frac{1}{a}, \frac{1}{b}, \frac{1}{c}}\rangle$.
Is this true? I think I may be able to use this conclusion to aid in my prime objective.
No, your assertion is not true: $\left<\vec{v},\vec{u}\right>$ is the sum of the products of their entries, i.e., $a\cdot(-1/a)+b\cdot(-1/b)+c\cdot(-1/c)=-3\neq 0$.
Trivially, the vector $\left<0,0,1\right>$ is orthogonal to your vector. A second vector can be found via using the cross product $\left<3,4,0\right>\times\left<0,0,1\right>$ and normalization. For references, see also this and this post.
Note that the two vectors you thus got span a two-dimensional space. Every vector in this two-dimensional space is orthogonal to your vector.
