How many orthogonal/perpendicular lines to this affine space

Question

I have an affine space in $V_6(\mathbb{R})$: $\{Y=(2,-2,0,1,-1,0)+a(1,-1,0,0,0,0)+b(1,0,0,1,-3,1)+c(1,-4,0,2,1,0)$ with $a,b,c \in \mathbb{R}\}$.

I have to find the number of:

1. orthogonal (to Y) lines passing through the origin
2. perpendicular (to Y) lines passing through the origin

Note that in the terminology I'm following two skew lines could be orthogonal but not perpendicular (they don't meet each other).

To find the orthogonal (to Y) lines I proceeded as follows: they have to be contained within $Y$'s orthogonal complement.

$Y^ \perp=((1,-1,0,0,0,0),(1,0,0,1,-3,1),(1,-4,0,2,1,0))^\perp$.

It's a complement, so its dimension is $6-3=3$.

$Y^\perp$ contains an infinite number of lines, so the first answer is infinite.

To find the perpendicular (to Y) lines I proceeded as follows:

$X$ is a generic line, spanned by the vector $(x_1,x_2,x_3,x_4,x_5,x_6)$: $$X=d(x_1,x_2,x_3,x_4,x_5,x_6)$$ where $d \in \mathbb{R}$.

If it's perpendicular to $Y$ then $<X,Y>=0$. So: $$dx_1(2+a+b+c)+dx_2(-2-a-4c)+dx_4(1+b+2c)+dx_5(-1-3b+c)+dx_6=0$$.

It's clear that there isn't only one solution, but I have to find the number of non-proportional 6-tuples $(x_1,x_2,x_3,x_4,x_5,x_6)$ that verify the equation.

Let's get one of these tuples: $(1,1,1,1,1,1)$ verifies the equation, but so does $(1,1,2,1,1,1)$, $(1,1,20000,1,1,1)$, $(1,1,a,1,1,1)$ with $a \in \mathbb{R}$ (in fact, $x_3$ doesn't appear in the eq). Furthermore they are non-proportional, so I've just found infinite (answer n.2) lines perpendicular to $Y$.

I'm not sure if it's right, if I have already imposed the incidence between the line and the space, if the second answer is possible and finally how should I calculate the distance between the space and the origin (on which perpendicular lines should I measure the distance?).

I'll be very grateful if you review what I wrote!

Update. I'll use the method I described in the comments to A.P.'s answer to find the distance between $Y$ and the origin.

A generic vector of $Y$ is something like this :$P+rA+sB+tC= \[ \left( \begin{array}{c} 2+r+s+t \\ -2-r-4t \\ 0 \\ 1+s+2t \\ -1-3s+t \\ 0+s \end{array} \right)\]$

The orthogonal complement $Y'= 0 + \{ (1,-1,0,0,0,0),(1,0,0,1,-3,1),(1,-4,0,2,1,0) \}^\perp $ passes through the origin.

Its carthesian equations representation is given by $\begin{equation} \begin{cases} x_1-x_2=0\\x_1+x_4-3x_5+x_6=0\\x_1-4x_2+2x_4+x_5=0 \end{cases} \end{equation}$

(from this representation I can check again that $0 \in Y'$).

Now, if I substitute $x_1, x_2$ ... with the coordinates of the generic vector of $Y$ I should get the intersection between $Y$ and the orthogonal complement passing through the origin and this point should be good to compute the distance. The problem is: the solution of the system is r = -132/103, s = -81/206, t = -43/206 [See Wolfram].

This solution is corresponding to the vector $(12/103, 12/103, 0, 39/206, -3/103, -81/206)$ whose length is $3 \sqrt{5/206}$ that is very different from A.P.'s elegant-way result. Where's the problem?

Answer 1

Edit So I understood your question as $V$ being a one-dimensional space $(a,b,c)$ fixed. It turned out you meant $V=\{... \text{ with } a,b,c\in\mathbb R\}$. This makes quite some change, however I will not delete my answer yet as I think it might help all the same.

When $(a,b,c)$ is fixed, $Y$ is a vector in $\mathbb R^6$ and $Y^\perp$ is a hyperplane, i.e. of dimension 5. Now, in the affine space, you want to translate this hyperplane so that it contains the origin. To do this, you can translate it along the vector $Y$. It doesn't change its dimension as a vector space, so for you first question, the orthogonal vector space is of dimension 5 (and of course it gives an infinity of solutions).

For the second question, the set of perpendicular lines is a subset of the orthogonal vector space. This subset consists in the set of lines of $Y^\perp$ passing through $0$ and the span of $Y$. But the span of $Y$ intersects $Y^\perp$ only at one point. So there is only one element of $Y^\perp$ which is perpendicular to $Y$.

Answer 2

Let $X$ be an affine sub-space of dimension $d$ in $\Bbb{A}^n(\Bbb{R})$, with directing vector space $V$. Then observe that any line through the origin can be parametrised as $\{t\vec{w} : t\in\Bbb{R}\}$ for some vector $\vec{w} \in \Bbb{R}^n$. Since every line orthogonal to $X$ is directed as a vector in $V^\perp$, it follows that there is a bijection between the points of the projective space $$ \Bbb{P}(V^\perp) $$ and the lines in $\Bbb{A}^n(\Bbb{R})$ through the origin orthogonal to $X$. Finally, note that $\Bbb{P}(V^\perp)$ has dimension $$ \dim_\Bbb{R} V^\perp - 1 = \operatorname{codim}_\Bbb{R} V - 1 = n - d - 1 $$ thus the only affine spaces with a finite number of orthogonal lines through the origin are $\Bbb{A}^n(\Bbb{R})$, which has none, and the hyperplanes, which have one.

Extra: If you've never seen a projective space, here's a nice way to think of $\Bbb{P}(V^\perp)$:

Consider the natural affine space structure $\Bbb{A}(V^\perp)$ on $V^\perp$. Then the lines through the origin of $\Bbb{A}^n(\Bbb{R})$ that are orthogonal to $V$ correspond $1:1$ to the lines through the origin in $\Bbb{A}(V^\perp)$ (make sure you understand this).

Now look at the unit $(n-d-1)$-sphere $S$ in $V^\perp$. Clearly every line through the origin intersects $S$ in two antipodal points. Conversely, every pair of antipodal points on $S$ defines a line through the origin.

Aha! But we can do better than this. Divide $S$ in half with a hyperplane $H$ through the origin, and call $T$ one half-sphere (border included). Then to every line through the origin and not in $H$ corresponds precisely to one point in the interior of $T$, while each of the other lines correspond to a pair of antipodal points on the border of $T$.

If $\sim$ is the equivalence relation on $\partial T$ defined as $$ \vec{x} \sim \vec{y} \iff \vec{x} \text{ is antipodal to } \vec{y} $$ then the quotient topological space $T/\sim$ is a concrete realization of $\Bbb{P}(V^\perp)$, and its dimension is defined as the dimension of the interior of $T$ as a hypersurface in $\Bbb{A}(V^\perp)$.

Exercise: Draw a picture and go through this construction for $\Bbb{P}(\Bbb{R}^2)$. Then try to do the same for $\Bbb{P}(\Bbb{R}^3)$. This should give you a good idea of what's going on here, and it should help you understand why I defined the dimension in that way.

Remark 1: Another way to construct $\Bbb{P}(V^\perp)$ is as the quotient (topological) space $(\Bbb{A}(V^\perp) \setminus \{\vec{0}\})/\sim$, where $$ \vec{x} \sim \vec{y} \iff \vec{x} = \lambda \vec{y} \text{ for some } \lambda \in \Bbb{R} \setminus \{0\} $$ i.e. as the space where proportional points of $\Bbb{A}(V^\perp)$ are identified. This is usually more practical to work with, but in this case I prefer the geometric picture of the other construction. By the way, proving that these two constructions are equivalent is a nice exercise.

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What about perpendicular lines, though? Those are just orthogonal lines that intersect $X$. So...

Fix a point $\vec{x} \in X$, let $\vec{v}_1,\dotsc,\vec{v}_d$ be a basis for $V$, and let $\vec{w}_{d+1},\dotsc,\vec{w}_n$ be a basis for $V^\perp$. Then a line $\ell \colon t\vec{w}$ is perpendicular to $X$ if and only if $\vec{w} = r_{d+1} \vec{w}_{d+1} + \dotsc + r_n \vec{w}_n$ and the linear system $$ \vec{x} + a_1 \vec{v}_1 + \dotsc + a_d \vec{v}_d = t \vec{w} \tag{1} \label{eq:1} $$ has a solution in $a_1,\dotsc,a_d,t$.

Clearly if $\vec{0} \in X$ every orthogonal line through the origin will also be perpendicular, but what if $\vec{0} \notin X$? Then $\eqref{eq:1}$ becomes $$ a_1 \vec{v}_1 + \dotsc + a_d \vec{v}_d + a_{d+1} \vec{w}_{d+1} + \dotsc + a_n \vec{w}_n = -\vec{x} \tag{2} \label{eq:2} $$ where $a_j = tr_j$ for $d+1 \leq j \leq n$. But this linear system has exactly one solution in $a_1,\dotsc,a_n$, because by construction $\vec{v}_1,\dotsc,\vec{v}_d,\vec{w}_{d+1},\dotsc,\vec{w}_n$ are linearly independent.

Remark 2: You interpret $\eqref{eq:1}$ in this way: given a point $\vec{x}$ on $X$ and a line $\ell$ through the origin, then $\ell$ intersects $X$ if and only if we can go from $\vec{0}$ to $\vec{x}$ by "walking" along $\ell$ for a while and then along a straight line parallel to (actually, contained in) $X$.

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Update: So, suppose that $\vec{0} \notin X$ and you want to compute the distance $\delta$ of $X$ from $\vec{0}$. First, observe that $\delta$ is the same as the length of the segment joining $\vec{0}$ and $\vec{\alpha} = X \cap \ell$, where $\ell$ is the unique line through $\vec{0}$ and perpendicular to $X$. In other words, identifying the point $\vec{\alpha}$ with the vector $\vec{\alpha}-\vec{0}$, we see that $\delta$ is just $\|\vec{\alpha}\|$.

Given the previous discussion, the naive way to compute $\delta$ is to find $\alpha$ by solving the linear system $\eqref{eq:2}$, and then computing $\|\alpha\|$... but we can do better — or, rather, the same but without having to compute $\vec{w}_{d+1},\dotsc,\vec{w}_n$.

Indeed, the "walk" I mentioned in remark 2 is nothing more than a decomposition of (the vector) $\vec{x}$ as a sum of two vectors: $\vec{x}_\perp$, perpendicular to $X$, and $\vec{x}_\parallel$, parallel to $X$. The uniqueness of $\ell$ means that $\vec{x}_\perp$ is the same for every $\vec{x} \in X$, so just pick one and compute: $$ \vec{x}_\perp = \vec{x} - (\vec{x} \cdot \vec{v}_1) \, \frac{\vec{v}_1}{\|\vec{v_1}\|^2} - \dotsb - (\vec{x} \cdot \vec{v}_d) \, \frac{\vec{v}_d}{\|\vec{v_d}\|^2}. $$ Important: As amd pointed out in his answer to a follow-up question, this formula is valid only if $\vec{v}_1,\dotsc,\vec{v}_d$ are pair-wise orthogonal. Indeed, otherwise suppose wlog that $\vec{v}_1 \cdot \vec{v}_2 \neq 0$; then we would end up subtracting from $\vec{x}$ at least $\sigma>1$ times the component of $\vec{x}$ along $\vec{v_1}$. This isn't a deal-breaker: we can easily extract an orthogonal — even orthonormal — basis from $\vec{v}_1,\dotsc,\vec{v}_d$, e.g. with the Gram-Schmidt process.

Where does this formula come from? We simply use the following three facts: $\vec{v}_1,\dotsc,\vec{v}_d$ form a basis for $V$; $\vec{x} = \vec{x}_\perp + \vec{x}_\parallel$; and for every $1 \leq i \leq d$ $$ \vec{x} \cdot \frac{\vec{v}_i}{\|\vec{v}_i\|} $$ is the component of $\vec{x}$ in the direction of $\vec{v}_i$.

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In your case $n = 6$ and $d = 3$, so there indeed infinitely many orthogonal lines through the origin. On the other hand, you can check that $\vec{0} \notin Y$, so there is exactly one perpendicular line through the origin.