While studying orthogonality and orthonormal bases, I found it hard to conceptualize orthogonal vectors in a Vector Space over $\Bbb R^n$ with $n>3$. In $\Bbb R^3$, we get the set of orthogonal vectors similar to the x,y,z axes.
In $\Bbb R^3$, vectors are orthogonal to planes, computed by looking at the vector's projection onto a plane(s). However, in $\Bbb R^4$, we are no longer looking at the projection onto a plane but rather the projection onto a vector space in $\Bbb R^3$?
So to find orthongonal vectors you would need to find the projection onto the span of three vectors in $\Bbb R^4$?
Rule is that the sum of dimensions of orthogonal subspaces is the dimension of the ambient space. So since in $\mathbb{R}^3$ you have 3 dimensions, the orthogonal space of a vector (which is a 1 dimensional object) is a plane, because 3-1 = 2 and the plane is 2 dimensional . So any object inside the plane orthogonal to that vector is orthogonal to the vector. Vectors in $\mathbb{R}^3$ can be orthogonal to other vectors, to lines, and ultimately to planes.
(Example: if I consider the vector $a = (0,0,1)$ his orthogonal plane is $\pi = \{z = 0\}$ which consists of the vectors $(x,y,0)$ with random $x$ and $y$. Hence the plane $\pi$ is orthogonal to $a$, and any vector inside $\pi$ is also orthogonal to $a$.)
Hence in $\mathbb{R}^4$, since you have 4 dimensions, the space orthogonal to a vector is a 3 dimensional space. (Ex:vector $(0,0,0,1)$ and space $(x,y,z,0)$ are orthogonal.
Yes, in $\mathbb{R}^4$ to find a vector orthogonal to a 3 dimensional space, you have to see if the projection of that vector on the span of 3 vectors generating your space is 0.