I’m a little bit stumped on the following problem.
Write down a vector equation for the cylinder in 3-dimensional space of radius r around the line described by $$\{ ( t, 2t, 3t ): t \in \Bbb R \}$$
Could anyone point out what I should do? I’m thinking about finding end-points using parametrics—something like $$ x = r \cos \theta, y=r \cos \theta, z=z $$
But I’m not sure. Thanks! (:
On
You’re starting off in the right direction by thinking about tilting the cylinder $x^2+y^2=r^2$ so that its axis lines up with the direction vector $(1,2,3)$ of the given line. A vector parameterization of this equation is $$r\cos t\,(1,0,0)+r\sin t\,(0,1,0)+s\,(0,0,1) \tag 1$$ so one might expect the tilted version to also have the form $$r\cos t\,\mathbf u+r\sin t\,\mathbf v+s\,\mathbf w. \tag 2$$ Obviously we can take $\mathbf w=(1,2,3)$ (we don’t care about the change of scale in this direction), but what to use for the other two? For a fixed $s$, we want (2) to represent a cross-section of the cylinder taken perpendicular to its axis, so $\mathbf u$ and $\mathbf v$ must be orthogonal to $\mathbf w$. In order to avoid distortions and get a circle of the correct size as this cross-section, they must be unit vectors. So, $\mathbf u$ can be any convenient unit vector that’s orthogonal to $\mathbf w$ and $\mathbf v = {1\over\|\mathbf w\|}\mathbf w\times\mathbf u$ also satisfies the criteria. These vectors form an orthogonal basis aligned with the cylinder.
For any nonzero vector $(a,b,c)$, at least two of $(0,c,-b)$, $(-c,0,a)$ and $(b,-a,0)$, which are all orthogonal to $(a,b,c)$, are also nonzero. For our $\mathbf w$, the choices are $(0,3,-2)$, $(-3,0,1)$ and $(2,-1,0)$. None of these result in particularly nice-looking unit vectors, so let’s just take $\mathbf u = \frac1{\sqrt{10}}(-3,0,1)$, giving $\mathbf v = \frac1{\sqrt{35}}(1,-5,3)$. A possible vector parameterization of the cylinder is therefore $$\frac1{\sqrt{10}}r\cos t\,(-3,0,1)+\frac1{\sqrt{35}}r\sin t\,(1,-5,3)+s\,(1,2,3). \tag 3$$
As a sanity check, an implicit Cartesian equation for the cylinder can be obtained in a similar fashion. Reflecting in the angle bisector of the $z$-axis and the given line will tilt the reference cylinder’s axis in the right way. We can use a reflection because of the cylinder’s symmetry, and I find a reflection more convenient than a rotation for this because the reflection matrix is both symmetric and is its own inverse, so I don’t have to remember which combination of inverse and transpose is required to transform the conic. As well, I don’t have to make any decisions about other degrees of freedom as I would for a rotation. Leaving out the details of this computation, the resulting equation is $$13x^2+10y^2+5z^2-4xy-3xz-6yz=14r^2.$$ The parameterization obtained above does indeed satisfy this equation.
As a comment from @Exodd suggests, you can establish a coordinate equation by computing the distance from the line, this distance is $r$ for all points on the cylinder.
A vectorial approach is as follows. Let $\vec u=(1,2,3)$ be the direction of the cylinder and let $\vec v$ be a unitary vector, perpendicular to $u$. Then every vector $\vec z$ pointing a point on the cylinder can be relied to the origin by the equations $$\vec z = t\vec u + r\vec v$$ $$\vec u\cdot \vec v=0,\; ||\vec v||=1$$
Note also that $\langle \vec u\rangle^\perp$ is the plane generated by $(-2,1,0),(-3,0,1)$. If you give an orthonormal base instead, you can write in a single equation what you want.