I am trying to solve a problem related to quadric surface(cylinder). The problem deals in finding the equation of a right circular cylinder whose guiding curve is a circle which passes through three points, (1,0,0), (0,1,0), (0,0,1).
Now, i know that there is an equation to calculate the equation of a right circular cylinder:
Eq: $(x-x_1)^2+(y-y_1)^2+(z-z_1)^2-\frac{\{l(x-x_1)+m(y-y_1)+n(z-z_1)\}^2}{l^2+m^2+n^2}=a^2$ (eq. i)
Now, i know that $(x_1,y_1,z_1)$ are the points through which the axial line of the cylinder passes, and $l,m,n$ are the direction ratios of the axial line.
Now, i think, if the three points on the circle is (1,0,0), (0,1,0), (0,0,1) , then the coordinates of the center of the circle must be (0,0,0), and since the axial line passes through the center of the circle, hence, $x_1,y_1,z_1 \;must \;be \;equal\; to\; 0$.
Now the symmetric form of the axial line will be: $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$, if so, the i have seen some exercises in the book, where they use 'l' in place of 'x', likewise , 'm' in place of 'y', if it is so, then if we use $x_1=y_1=z_1=0$ and $x=l,y=m,z=n$ in equation (i) above, then it shows error.
This is the problem I am dealing with. Can anybody help me? (actually, this is a proving question , where the required equation of the cylinder is $x^2+y^2+z^2-yz-zx-xy=1$)
Hint.
Calling
$$ p_1 = (1,0,0)\\ p_2 = (0,1,0)\\ p_3 = (0,0,1)\\ \vec n = (p_1-p_2)\times(p_3-p_2)\\ p_0 = (x_0,y_0,z_0) $$
with $\vec n$ normal to the circle and $p_0$ representing the circle's center we have
$$ \cases{ (p_1-p_0)\cdot (p_1-p_0) = r^2\\ (p_2-p_0)\cdot (p_2-p_0) = r^2\\ (p_3-p_0)\cdot (p_3-p_0) = r^2\\ (p_1-p_0)\cdot \vec n = 0 } $$
Solving those equations for $p_0, r$ we obtain the circle's center and radius. Once you have $p_0, r$ you can use the equation (eq.i)