Obtain equation of right circular cylinder on the circle through points $(a,0,0),(0,b,0)$ and $(0,0,c)$ as the guiding curve.
I proceeded by considering this circle as intersection of sphere from the given points along with origin and the plane $x/a+y/b+z/c=1$ and obtained the equation of axis of the cylinder as it is supposed to pass through the centre of sphere and is perpendicular to the plane. But can not proceed after that as need to calculate the radius of the circle through these points and unable to get the radius.
Is the approach wrong or needs some change? Any help appreciated
$\endgroup$3 Answers
$\begingroup$Let's assume the center of the circle that passes through $3$ given points is $(x_0, y_0, z_0).$
One of the ways to do it -
- find the equation of the plane through these $3$ points which you have already done.
- Equate the distance between the center and the $3$ points on the circle (as they are radii of the circle).
- As the center also lines on the plane, we get one more equation.
Solving these equations gives you the center of the circle and then distance to any of the $3$ points is the radius.
Axis of the cylinder is the line orthogonal to the plane and passing through the center of the circle that we just obtained.
But I find it faster to use barycentric coordinates. Please see below.
Our circle is going through points $A(a,0,0), B(0,b,0), C(0,0,c)$ and it is the circumcircle of the $\triangle ABC$.
So coordinates of the center $O$ is given by $\alpha (a,0,0) + \beta (0,b,0) + \gamma (0,0,c)$
where $\alpha + \beta + \gamma = 1$
The ratio of $\alpha, \beta, \gamma$ is given by the sides of the triangle as follows -
$\alpha : \beta : \gamma = BC^2(AB^2+CA^2-BC^2):CA^2(AB^2+BC^2-CA^2):AB^2(BC^2+CA^2-AB^2)$
and you get $ \, \alpha : \beta : \gamma = a^2(b^2+c^2):b^2(c^2+a^2):c^2(a^2+b^2)$
Earlier we said coordinates of the center $O = \alpha (a,0,0) + \beta (0,b,0) + \gamma (0,0,c)$
where $\alpha + \beta + \gamma = 1$
So, $O = \displaystyle (\frac{a^3(b^2+c^2)}{2(a^2b^2+b^2c^2+c^2a^2)}, \frac{b^3(c^2+a^2)}{2(a^2b^2+b^2c^2+c^2a^2)}, \frac{c^3(a^2+b^2)}{2(a^2b^2+b^2c^2+c^2a^2)})$
Distance from $O$ to any of the points $A, B, C$ will give you the radius.
You can also find the parametric equation of the axis of the cylinder as it is orthogonal to the plane with points $A, B$ and $C$ and it passes through point $O$.
$\endgroup$ $\begingroup$A point in the plane has the coordinates $u(a,0,0)+v(0,b,0)+w(0,0,c)=(ua,vb,wc)$ with $u+v+w=1$. Let us express that the center is in the plane and is equidistant from the given points:
$$(ua-a)^2+(vb)^2+(wc)^2=(ua)^2+(vb-b)^2+(wc)^2=(ua)^2+(vb)^2+(wc-c)^2$$
or after expansion and simplification,
$$a^2(1-2u)=b^2(1-2v)=c^2(1-2w).$$
Let this common value be $d^2$. Then
$${d^2}\left(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\right)=3-2=1$$ and we obtain the coordinates of the center:
$$ua=\frac a2-\frac{d^2}{2a},vb=\frac b2-\frac{d^2}{2b},wc=\frac c2-\frac {d^2}{2c}.$$
And the direction of the axis is $\left(\dfrac1a,\dfrac1b,\dfrac1c\right)$. The unit vector in the direction of the axis is $\vec t=d\left(\dfrac1a,\dfrac1b,\dfrac1c\right)$.
The implicit equation of the cylinder expresses that the distance of a point to the axis is the radius, i.e.
$$\|(x-ua,y-vb,z-wc)\times\vec t\|=r.$$
$\endgroup$ $\begingroup$In general, the parametric equations of a cylinder are of the type:
$$ (x,\,y,\,z) = (x(u),\,y(u),\,z(u)) + (l,\,m,\,n)\,v \; \; \; \text{for} \; (u,\,v) \in I \times J \subseteq \mathbb{R}^2\,, $$
where $(x,\,y,\,z) = (x(u),\,y(u),\,z(u))$ for $u \in I$ and $(l,\,m,\,n) \ne (0,\,0,\,0)$ are respectively the directrix curve and the direction of the generating lines of the cylinder. In particular, in order for this cylinder to be right circular, the directrix curve must be a circle with center $C(x_C,\,y_C,\,z_C)$ and radius $r > 0$ lying in the plane $\alpha$ whose direction is $(l,\,m,\,n) \ne (0,\,0,\,0)$.
Specifically, the plane passing through three non-aligned points $P_1(x_1,\,y_1,\,z_1)$, $P_2(x_2,\,y_2,\,z_2)$, $P_3(x_3,\,y_3,\,z_3)$ has a director vector:
$$ (l,\,m,\,n) = \left(P_2 - P_1\right) \land \left(P_3 - P_1\right) $$
and indicating with $P(x,\,y,\,z)$ its generic point, this plane has a Cartesian equation:
$$ \alpha : \; (l,\,m,\,n) \cdot \left(P - P_1\right) = 0\,. $$
Thus, solving the system of linear equations:
$$ \begin{cases} ||C - P_1||^2 = ||C - P_2||^2 \\ ||C - P_2||^2 = ||C - P_3||^2 \\ (l,\,m,\,n) \cdot \left(C - P_1\right) = 0 \end{cases} $$
it's possible to determine the three coordinates of $C$ and consequently the radius of the circle $r = ||C - P_1|| = ||C - P_2|| = ||C - P_3||$. In particular, if we have $P_1(a,\,0,\,0)$, $P_2(0,\,b,\,0)$, $P_3(0,\,0,\,c)$, we obtain respectively:
$$ (l,\,m,\,n) = (b\,c,\,a\,c,\,a\,b) \,; \\ C\left(\frac{a^3\left(b^2+c^2\right)}{2\,a^2\left(b^2+c^2\right)+2\,b^2\,c^2},\,\frac{b^3\left(a^2+c^2\right)}{2\,a^2\left(b^2+c^2\right)+2\,b^2\,c^2},\,\frac{c^3\left(a^2+b^2\right)}{2\,a^2\left(b^2+c^2\right)+2\,b^2\,c^2}\right) \,; \\ r = \sqrt{\frac{\left(a^2+b^2\right)\left(a^2+c^2\right)\left(b^2+c^2\right)}{4\,a^2\left(b^2+c^2\right) + 4\,b^2\,c^2}} \,. $$
We therefore have all the ingredients to determine the parametric equations of the circle understood as the intersection between the sphere of center $C$ and radius $r > 0$ with $\alpha$:
- if $m^2+n^2 \ne 0$, we have:
$$ \begin{cases} x(u) = x_C + r\left(\frac{\sqrt{m^2+n^2}}{\sqrt{l^2+m^2+n^2}}\,\cos u\right) \\ y(u) = y_C + r\left(\frac{-l\,m}{\sqrt{m^2+n^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u + \frac{n}{\sqrt{m^2+n^2}}\,\sin u\right) \\ z(u) = z_C + r\left(\frac{-l\,n}{\sqrt{m^2+n^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u - \frac{m}{\sqrt{m^2+n^2}}\,\sin u\right) \\ \end{cases} \; \; \; \text{for} \; u \in [0,\,2\pi) \,; $$
- if $l^2+n^2 \ne 0$, we have:
$$ \begin{cases} x(u) = x_C + r\left(\frac{-l\,m}{\sqrt{l^2+n^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u + \frac{n}{\sqrt{l^2+n^2}}\,\sin u\right) \\ y(u) = y_C + r\left(\frac{\sqrt{l^2+n^2}}{\sqrt{l^2+m^2+n^2}}\,\cos u\right) \\ z(u) = z_C + r\left(\frac{-m\,n}{\sqrt{l^2+n^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u - \frac{l}{\sqrt{l^2+n^2}}\,\sin u\right) \\ \end{cases} \; \; \; \text{for} \; u \in [0,\,2\pi) \,; $$
- if $l^2+m^2 \ne 0$, we have:
$$ \begin{cases} x(u) = x_C + r\left(\frac{-l\,n}{\sqrt{l^2+m^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u + \frac{m}{\sqrt{l^2+m^2}}\,\sin u\right) \\ y(u) = y_C + r\left(\frac{-m\,n}{\sqrt{l^2+m^2}\,\sqrt{l^2+m^2+n^2}}\,\cos u - \frac{l}{\sqrt{l^2+m^2}}\,\sin u\right) \\ z(u) = z_C + r\left(\frac{\sqrt{l^2+m^2}}{\sqrt{l^2+m^2+n^2}}\,\cos u\right) \\ \end{cases} \; \; \; \text{for} \; u \in [0,\,2\pi) \,; $$
which, respectively, substituted in the initial parametric equations give what is desired.
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