Answer to Question #91550 in Analytic Geometry for Ra

Question

Find the equation of the cyclinder with base x²+y²+z²+3x+3y-z=1, x+y+2z=2.

Answer 1

Given plane equation is (x+y+2z-2)=0, sphere equation is (x²+y²+z²+3x+3y-z)=1

$x^2+2\cdot{3 \over 2}x+{9 \over 4}-{9 \over 4}+y^2+2\cdot{3 \over 2}y+{9 \over 4}-{9 \over 4}+$

$+z^2-2\cdot{1 \over 2}z+{1 \over 4}-{1 \over 4}=1$

$\big(x+{3 \over 2}\big)^2+\big(y+{3 \over 2}\big)^2+\big(z-{1 \over 2}\big)^2={23 \over 4}$

$Centre (-u,-v,-w)=\big(-{3 \over 2},-{3 \over 2},{1 \over 2}\big)$

$Radius\ \ r=\sqrt{{23 \over 4}}={\sqrt{23} \over 2}$

Length of the perpendicular from $\big(-\dfrac{3}{2},-\dfrac{3}{2},\dfrac{1}{2}\big)$ to the plane (x+y+2z-2) =0 is 

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

Here, $a=1,b=1,c=2,d=-2,x_1=-\dfrac{3}{2},y_1=-\dfrac{3}{2},z_1=\dfrac{1}{2}$

$OM=\dfrac{\big|-\dfrac{3}{2}-\dfrac{3}{2}+2\cdot\dfrac{1}{2}-2\big|}{\sqrt{1^2+1^2+2^2}}=\dfrac{2\sqrt6}{3}$

$AM^2=OA^2-OM^2=({\sqrt{23} \over 2})^2-({2\sqrt6 \over 3})^2={37 \over 12}$

The equation of the axis of the cylinder is 

${x-\alpha \over l}={y-\beta \over m}={z-\gamma \over n}$

Here, $\alpha=-\dfrac{3}{2},\beta=-\dfrac{3}{2},\gamma=\dfrac{1}{2}, l=1,m=1,n=2.$

${x+\dfrac{3}{2} \over 1}={y+\dfrac{3}{2} \over 1}={z-\dfrac{1}{2} \over 2}$

$l^2+m^2+n^2=6$

$Radius\ \ r^*=AM=\sqrt{{37 \over 12}}={\sqrt{111} \over 6}$

The equation of Right Circular Cylinder is, 

$[n(y-\beta)-m(z-\gamma)]^2+[l(z-\gamma)-n(x-\alpha)]^2+$ $+[m(x-\alpha)-l(y-\beta)]^2=(r^*)^2(l^2+m^2+n^2)$

$[2(y+{3 \over 2})-1(z-{1\over 2})]^2+[1(z-{1 \over 2})-2(x+{3\over 2})]^2+$ $+[1(x+{3 \over 2})-1(y+{3 \over 2})]^2=({\sqrt{111} \over 6})^2(6)$

$4y^2+z^2+{49 \over 4}-4yz+14y-7z+z^2+4x^2+{49 \over 4}-$ $-4xz-7z+14x+x^2-2xy+y^2={37 \over 2}$

$5x^2+5y^2+2z^2-2xy-4xz-4yz+14x+14y-14z+6=0$