If I have a vector $(0,4,4)$ and was finding perpendicular unit vectors to this, is it the same case as finding perpendicular unit vectors for the vector $(4,4)$? Meaning a maximum of two possible perpendicular unit vectors?
When I draw and visualize the space that $\Bbb R^3$ vectors occupy, it would seem logical to me that if one component is equal to zero then the vector could be treated as if it was in $\Bbb R^2$ for its non-zero components.
On
Not really. In $\Bbb R^3$ you can obtain many more orthogonal vectors to $\vec v$ since you can consider one and make it rotate around $\vec v$, keeping the plane where it lies fixed. Thus a vector has a "plane" of orthogonal vectors, while the vector in $\Bbb R^2$ will only have a line (which is basically the projection of the said plane over a modified "$\Bbb R^2$")
On
There will be exactly two perpendicular unit vectors which also have the first coordinate zero, i.e., that also lie in the plane $x=0$. There are infinitely many unit vectors perpendicular to $(0,4,4)$ if you allow vectors from all of $\mathbb R^3$.
On
In this situation, no!
$(0,4,4)$ have more than two perpendicular unit vectors. For instance, $(1,0,0)$, $\frac{\sqrt{2}}{2}(0,1,-1)$ and $\frac{\sqrt{2}}{2}(0,-1,1)$.
On
In the plane, where $x = 0$, there are exactly two unit vectors perpendicular to $(0, 4, 4)$, namely $$ \begin{align} \tfrac{1}{\sqrt{2}}(0, 1, -1) &= \left( 0, \tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}} \right) \\ \text{and} \quad \tfrac{1}{\sqrt{2}}(0, -1, 1) &= \left( 0, -\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}} \right). \end{align} $$
However, in $\mathbb{R}^3$, there is an entire circle's worth of unit vectors perpendicular to $(0, 4, 4)$.
Consider the two unit vectors $\hat{v} = (1, 0, 0)$ and $\hat{w} = \left( 0, \tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}} \right)$. You can parametrize all the unit vectors perpendicular to $(0, 4, 4)$ in terms of these two: $$ \hat{x}_\theta = (\cos \theta) \hat{v} + (\sin \theta) \hat{u} = \left( \cos \theta, \frac{\sin \theta}{\sqrt{2}}, -\frac{\sin \theta}{\sqrt{2}} \right). $$
Any nonzero vector in $\mathbb{R}^3$ will have an orthonormal set spanned by two vectors.
If you ignore the zero component and draw it in $\mathbb{R}^2$, the two basis vectors for the orthonormal set will be one on the paper (orthogonal in the $\mathbb{R}^2$plane) and another vector "coming out of the paper," orthonormal to the first.