Answer to Question #293618 in Analytic Geometry for Elley

Question

Find the equation of the cylinder with base 𝑥² + 𝑦² + 𝑧² − 3𝑥 − 6𝑧 + 9 = 0, 𝑥 − 2𝑦 + 2𝑧 − 6 = 0.

Answer 1

Given the plane:

$𝑥 − 2𝑦 + 2𝑧 − 6 = 0$

and the sphere equation:

$𝑥^2 + 𝑦^2 + 𝑧^2 − 3𝑥 − 6𝑧 + 9 = 0$

$\displaystyle \Rightarrow \left(x-\frac{3}{2}\right)^2+\left(y-0\right)^2+(z-3)^2=\left(\frac{3}{2}\right)^2$by completing the square method

Then the center of the sphere is $\displaystyle \left(\frac{3}{2},\ 0,\ 3\right)$ with radius $\displaystyle \frac{3}{2}.$

Now, the length of the perpendicular from $\displaystyle \left(\frac{3}{2},\ 0,\ 3\right)$ to the plane $𝑥 − 2𝑦 + 2𝑧 − 6 = 0$ is;

$\displaystyle OM=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$

Here, $\displaystyle a=1, b=-2, c=2, d=-6, x_1=\frac{3}{2}, y_1=0, z_1=3$. Thus,

$\displaystyle OM=\frac{|1\cdot \frac{3}{2}+(-2)\cdot 0+2\cdot 3+(-6)|}{\sqrt{1^2+(-2)^2+2^2}}=\frac{|\frac{3}{2}|}{3}=\frac{1}{2}$

Next, $\displaystyle AM^2=OA^2-OM^2=\left(\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=2$

The equation of the axis of the cylinder is;

$\displaystyle \frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{x-\gamma}{n}, \text{ Here, }(\alpha,\ \beta,\ \gamma)=\left(\frac{3}{2},\ 0,\ 3\right),\ l=1,\ m=-2,\ n=2$.

$\displaystyle \Rightarrow \frac{x-\frac{3}{2}}{1}=\frac{y-0}{-2}=\frac{x-3}{2}, \text{ and }l^2+m^2+n^2=9$

The equation of the Right Circular Cylinder is;

$\displaystyle [n(y-\beta)-m(z-\gamma)]^2+[l(z-\gamma)-n(x-\alpha)]^2+[m(x-\alpha)-l(y-\beta)]^2=AM^2(l^2+m^2+n^2)\\ \Rightarrow\\ \left[2(y-0)+2(z-3)\right]^2+\left[1(z-3)-2\left(x-\frac{3}{2}\right)\right]^2+\left[-2\left(x-\frac{3}{2}\right)-1(y-0)\right]^2=2(9)\\ \Rightarrow \left[2y+2z-6)\right]^2+\left[z-2x\right]^2+\left[-2x+3-y\right]^2=18\\ \Rightarrow 8x^2+5y^2+5z^2+8yz-4xz+4xy-12x-30y-24z+27=0$