Answer to Question #202767 in Analytic Geometry for tanya

Question

Find the equation of the cylinder with base

x² + y² + z² + 3x+ 3y−z = 1,x + y+ 2z = 2.

Answer 1

Solution:

Given

x² + y² + z² + 3x+ 3y−z = 1, x + y+ 2z = 2

$\Rightarrow (x-\frac32)^2-(\frac32)^2+(y-\frac32)^2-(\frac32)^2+(z-\frac12)^2-(\frac12)^2=1 \\ \Rightarrow (x-\frac32)^2+(y-\frac32)^2+(z-\frac12)^2=\frac{23}4$

$p=(x-\frac32, y-\frac32, z-\frac12) \\ \vec{n}=(1,1,2)$

$p_{1}=\left(0,0, 1\right)$

$S \rightarrow\ \left\langle|p\|^{2}-(\frac{23}4)^{2}\right\rangle=x^2 + y^2 + z^2 + 3x+ 3y−z = 1$

$\Pi \rightarrow\left\langle\vec{n}, p-p_{1}\right\rangle=x + y+ 2z = 2$

The line $L \rightarrow p_{0}+\lambda \vec{n}$

is potentially a cylinder generatrix, providing that $p_{0} \in S \cap \Pi$

then

$\left\|p_{0}+\lambda \vec{n}\right\|^{2}=\left\|p_{0}\right\|^{2}+2 \lambda\left\langle p_{0}, \vec{n}\right\rangle+\lambda^2\|\vec{n}\|^{2}=\frac{23}4$

$\Rightarrow \lambda^2\|\vec{n}\|^{2}+2 \lambda\left\langle p_{0}, \vec{n}\right\rangle+\left\|p_{0}\right\|^{2}-\frac{23}4=0$

which is a quadratic equation in variable $\lambda$.

Solving for $\lambda$

$\lambda=\dfrac{-\left\langle p_{0}, \vec{n}\right\rangle \pm \sqrt{\left\langle p_{0}, \vec{n}\right\rangle^{2}-4(\|\vec{n}\|^{2})(\left\|p_{0}\right\|^{2}-\frac{23}4})}{\|\vec{n}\|^{2}}$

The tangency condition requires that

${\left\langle p_{0}, \vec{n}\right\rangle^{2}-4(\|\vec{n}\|^{2})(\left\|p_{0}\right\|^{2}-\frac{23}4})=0$

$\Rightarrow {\left\langle p_{0}, \vec{n}\right\rangle^{2}-4\|\vec{n}\|^{2}\left\|p_{0}\right\|^{2}+23\|\vec{n}\|^{2}}=0$ ...(i)

We have $p_0=(x_0,y_0,z_0), \vec{n}=(1,1,2),\|\vec{n}\|^{2}=6,\|\vec{p_0}\|^{2}=x_0^2+y_0^2+z_0^2$

So, (i) becomes,

$(x_0+y_0+2z_0)^2-4(6)(x_0^2+y_0^2+z_0^2)+23(6)=0 \\\Rightarrow x_0^2+y_0^2+4z_0^2+2x_0y_0+4y_0z_0+4x_0z_0 -24x_0^2-24y_0^2-24z_0^2+138=0$

$\Rightarrow 2x_0y_0+4y_0z_0+4x_0z_0-23x_0^2-23y_0^2-20z_0^2 +138=0$

which gives the equation of cylinder.