Finding all the points whose distance from a given line equals 1

Question

I have a line in the vector form: $(x,y,z) = t(1,1,1) + (0,0,1)$ and I need to find a vector that represents all the points whose distance from this line equals $1$. I started calculating the distance between a point $(a,b,c)$ and this line using the vetorial product... and I got a big equation in terms of $a$, $b$ and $c$. I think this equation describes a cylinder whose centerline is $(x,y,z) = t(1,1,1) + (0,0,1)$. But I don't know how to get rid of this equation and obtain the vector...

Can anyone help me? Thank you

Answer 1

Actually, there are infinitely many points as 1 unit far from the line. For those points, namely having coordinates $(p,q,r)$, there exists a unit vector $(a,b,c)$ normal to the vector of line $(1,1,1)$ and a point $(x,y,z)$ on the line such that $$ (p,q,r)=(x,y,z)+(a,b,c). $$ The unicity of $(a,b,c)$, along its normality implies $$ {a^2+b^2+c^2=1 \\ (a,b,c)\cdot(1,1,1)=a+b+c=0, } $$ which yields $$ {c=-a-b \\ a^2+b^2+ab=\frac{1}{2}\implies \\ b=\frac{-a\pm\sqrt{2-3a^2}}{2} \\ c=\frac{-a\mp\sqrt{2-3a^2}}{2}. } $$Since the point $(x,y,z)$ is arbitrary, the equation of all points far from the line as much as 1 is $$ \left(t+a,t+\frac{-a\pm\sqrt{2-3a^2}}{2},1+t+\frac{-a\mp\sqrt{2-3a^2}}{2}\right) $$ for $|a|\le \sqrt\frac{2}{3}$.

Answer 2

As you said, the points satisfying the given condition lie on a cylinder whose axis is the given line $(0, 0, 1) + t (1, 1, 1) $

Let $r_0 = (0, 0, 1)$ , and $d = (1, 1, 1)$ and let $r = 1$ then the equation of this cylinder is

$(p - r_0)^T (I - \dfrac{dd^T}{d d^T} ) (p - r_0) = r^2 $

Where $p$ is any point on the cylinder.

Simplifying the above equation, it becomes

$ (p - (0, 0, 1) )^T Q (p - (0, 0, 1) ) = 1 $

where $ Q = \dfrac{1}{3} \begin{bmatrix}2&&-1&&-1\\-1&&2&&-1\\-1&&-1&&2\end{bmatrix}$

Expanding the quadratic form of the above equation gives us,

$ x^2 + y^2 + (z - 1)^2 - x y - x(z-1) - y(z-1) = \frac{3}{2} $

And this is the algebraic equation describing all the points on the cylinder.

But this not what you want. You want a vector equation of the cylinder, and for that,

we construct any two perpendicular unit vectors $u_1$ and $u_2$ which are perpendicular to the axis vector $(1,1,1)$. Two such vectors are $(1, -1, 0)/\sqrt{2}$ and $(1, 1, -2)/\sqrt{6} $, then the vector equation of all the points on the cylinder is:

$ r(\phi, s ) = (0, 0, 1) + (\cos \phi) u_1 + (\sin \phi) u_2 + s (1, 1, 1) $

where $\phi \in [0, 2 \pi) $ and $s \in \mathbb{R}$.