Find the equation of the cylinder whose two directions are given.

Question

The radial directions of a cylinder is given by $$ \begin{array} wx^2+y^2=5^2, \\ z=0 . \end{array} $$ and axial directions of a cylinder is $$\vec a=(5,3,2)$$ respectively. Find the equation of the cylinder.

Because of the lockdown due to the pandemic, I was not able to attend the lessons. Therefore, I'm having trouble solving this kind of exercises. I would kindly ask you to give me a solution and help me understand (so that I would be able to solve the rest problems myself). I've gained some background on cylinders, but that seems not to be sufficient to solve this.

Answer 1

At height $z$ the horizontal section of the cylinder is a circle, with center $\big({5\over2}z,{3\over2}z,z\big)$ and radius $5$. Hence the equation of the cylinder is: $$ \left(x-{5\over2}z\right)^2+\left(y-{3\over2}z\right)^2=5^2 $$

Answer 2

Calling $p = (x,y,z)$ and giving

$$ L\to p = p_0 +\lambda\vec a\\ C\to \left(p_0\cdot p_0\right)_{z=0} = r^2\\ $$

$C$ is the projection along the $\vec a$ direction onto the plane $z=0$ of the ellipsoid

$$ E\to x^2+y^2+c_{33}z^2+c_{13} x z+c_{23}yz=c_0\Rightarrow p_0\cdot M\cdot p_0=c_0 $$

so $L\cap E$ is given by

$$ (p-\lambda\vec a)\cdot M\cdot(p-\lambda\vec a) = c_0 $$

or

$$ \lambda^2\vec a\cdot M\cdot\vec a -2\lambda p\cdot M\vec a+p\cdot M\cdot p = c_0 $$

solving for $\lambda$ we have

$$ \lambda = p\cdot M\vec a\pm \sqrt{\left( p\cdot M\vec a\right)^2-(\vec a\cdot M\cdot\vec a)(p\cdot M\cdot p-c_0)} $$

but we need tangency between $E$ and $L$ hence

$$ \left( p\cdot M\vec a\right)^2-(\vec a\cdot M\cdot\vec a)(p\cdot M\cdot p-c_0)=0 $$

This is the cylinder equation. Additionally we know that for $z=0$ the intersection of $E$ with the plane $z=0$ is $C$ so equating coefficients

$$ \left(\left( p\cdot M\vec a\right)^2-(\vec a\cdot M\cdot\vec a)(p\cdot M\cdot p-c_0)\right)_{z=0}\equiv x^2+y^2-r^2 $$

we get

$$ c_0 = 25,\ c_{13} = -5,\ c_{33} = \frac{33}{4} $$

so the cylinder equation is

$$ x^2-5 x z+y^2-3 y z+\frac{17 z^2}{2}-25=0 $$

Follows an illustration.