Question #293618

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


1
Expert's answer
2022-02-10T03:39:58-0500

Given the plane:

𝑥2𝑦+2𝑧6=0𝑥 − 2𝑦 + 2𝑧 − 6 = 0

and the sphere equation:

𝑥2+𝑦2+𝑧23𝑥6𝑧+9=0𝑥^2 + 𝑦^2 + 𝑧^2 − 3𝑥 − 6𝑧 + 9 = 0

(x32)2+(y0)2+(z3)2=(32)2\displaystyle \Rightarrow \left(x-\frac{3}{2}\right)^2+\left(y-0\right)^2+(z-3)^2=\left(\frac{3}{2}\right)^2by completing the square method

Then the center of the sphere is (32, 0, 3)\displaystyle \left(\frac{3}{2},\ 0,\ 3\right) with radius 32.\displaystyle \frac{3}{2}.

Now, the length of the perpendicular from (32, 0, 3)\displaystyle \left(\frac{3}{2},\ 0,\ 3\right) to the plane 𝑥2𝑦+2𝑧6=0𝑥 − 2𝑦 + 2𝑧 − 6 = 0 is;

OM=ax1+by1+cz1+da2+b2+c2\displaystyle OM=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

Here, a=1,b=2,c=2,d=6,x1=32,y1=0,z1=3\displaystyle a=1, b=-2, c=2, d=-6, x_1=\frac{3}{2}, y_1=0, z_1=3. Thus,

OM=132+(2)0+23+(6)12+(2)2+22=323=12\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, AM2=OA2OM2=(32)2(12)2=2\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;

xαl=yβm=xγn, Here, (α, β, γ)=(32, 0, 3), l=1, m=2, n=2\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.

x321=y02=x32, and l2+m2+n2=9\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;

[n(yβ)m(zγ)]2+[l(zγ)n(xα)]2+[m(xα)l(yβ)]2=AM2(l2+m2+n2)[2(y0)+2(z3)]2+[1(z3)2(x32)]2+[2(x32)1(y0)]2=2(9)[2y+2z6)]2+[z2x]2+[2x+3y]2=188x2+5y2+5z2+8yz4xz+4xy12x30y24z+27=0\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


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