The question is below:
Let $a, b, c$ be nonzero real numbers and let $U$ be the line of equations $ax=by=cz$ (which means that all the points $(x,y,z$) on the line satisfy that equation). Find (describe it by an equation, or equations, in terms of $x,y,z$) the orthogonal complement of U with respect to the standard dot product, that is find the subspace $V$ of $\mathbb{R^3}$ such that all the vectors in $V$ are orthogonal to all the vectors in $U$ and $\mathbb{R^3}=U\oplus V$. Also what does V represent geometrically?
My idea to the problem is as follows: $$U=\{(t/a,t/b,t/c) :t \in \mathbb{R}\}$$
$$V=\{(x,y,z): (x,y,z).(t/a,t/b,t/c)=0\}$$
which yields $tx/a+ty/b+tz/c =0$
I really do not know how to explain why I chose these for $U$ and $V$. Also, I do not know what $V$ represent geometrically. Please, can you help me check my idea to the problem and correct me where I might have gone wrong in an explainable way? Also what does $V$ represent geometrically?
I will be grateful if you put me through. Thanks so much.
Intuitively, the orthogonal complement of $X$ is the set of vectors that are orthogonal to every vector in $X$. So, when you work in the vector space $\mathbb{R^3}$, and you have a line that goes through the origin ($1$-dimensional subspace), the orthogonal complement will have dimension $2$ (and therefore be a plane). This is easy to visualise: Draw a line in a $3$ dimensional coordinate system (through the origin). The plane that is perpendicular to this line and goes through the origin will be the orthogonal complement.
So now, I will give you a hint to continue:
Every plane can be written in the form:
$ax+by+cz + d = 0$
The vector $(a,b,c)$ is a normal vector on the plane, so you can use this information to find the desired plane.
EDIT:
Your method with the vector scalar product was correct
The only thing left to do is eliminitation of $t$, we find the plane:
$bcx + acy + abz = 0$ (multiply both sides with $abc/t$), which is the subspace we were looking for (no need for a determinant here)