I am trying to solve a problem where I need to find the equation of a right circular cylinder. The problem tells us that the guiding curve of the cylinder passes through $(1,0,0),(0,1,0),(0,0,1)$ and it mentions nothing more.
Now, I calculated the centre of the above circle through three mentioned points and found the center to be $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ and radius to be $r=\sqrt\frac{2}{3}.$
Now it is necessary to find the direction ratios of the axial line to calculate the equation of the right circular cylinder. It is obvious that the axial line passes through the center of the circle which is $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. But, I need one more point that the axial line passes through in order to find it's direction ratios. Then I plan to use the found direction ratios in the below equation to find the equation of the cylinder.
$(x-x_1)^2+(y-y_1)^2+(z-z_1)^2-\frac{\{l(x-x_1)+m(y-y_1)+n(z-z_1)\}^2}{l^2+m^2+n^2}=a^2$ (eq. i),
where, $l,m,n$ are the direction ratio's of the axial line,
$(x_1,y_1,z_1)$ are the points through which the axial line passes, which we can take as $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$,
$a$ is the radius of the circle(which is $\sqrt \frac{2}{3}$ )
To solve this question I need values of $l,m,n$ which I tried to find but could not. Can anyone guide me clearly?
Note that the points $A(1,0,0), B(0,1,0)$ and $C(0,0,1)$ form an equilateral triangle. The normal vector to the plane in which this triangle lies will possess your desired direction ratios.
First, find
$\vec{AB} = (-1, 1, 0)$
$\vec{BC} = (0, -1, 1)$
And so$$\vec{AB} \times \vec{BC} = (1, 1, 1) =(l,m,n)$$