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.


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)

  • $\begingroup$ Thanks for your response. Is my initially proposition for U and V correct? $\endgroup$ – Tony.... May 1 '17 at 20:53
  • $\begingroup$ I know to find the equation of the plane, i will use the determinant method but i do not know the points the plane contain. Can you explain to me in details? Thanks. $\endgroup$ – Tony.... May 1 '17 at 21:46
  • $\begingroup$ Will do once I get back from university. $\endgroup$ – user370967 May 2 '17 at 6:49
  • $\begingroup$ Okay. Thanks you so much. $\endgroup$ – Tony.... May 2 '17 at 11:23
  • $\begingroup$ As promised, I edited my post $\endgroup$ – user370967 May 2 '17 at 19:08

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