In a three-dimensional Cartesian coordinate system, consider parameter $m$ and plane $$(P_m) \colon (m - 1)x + (2 - m)y - mz + (2m - 1) = 0$$ What are the coordinates of the centre of the fixed circle on which the projection of point $A(1; 0; 3)$ on plane $(P_m)$ runs?
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
This was actually from my mid-semester exam, it was a 90-minute nightmare in broad daylight. How many times did I hug my best friend who was sitting next to me out of hopelessness again?
Here is my attempt of solving this problem, with the assistance of my trusty laptop.
It is evident that the vector perpendicular to plane $(P_m)$ is $\vec{n} = (m - 1; 2 - m; 2m - 1)$. Let $M$ be the projection of point $A$ on plane $(P_m)$.
Since $\overrightarrow{AM} \parallel \vec{n}$, the coordinates of point $M$ could be written in the form of $(mn - (n - 1); 2n - mn; 2mn - (n - 3))$, with $n$ being another parameter.
Furthermore, as $M$ lies on plane $(P_m)$, $$(m - 1)[mn - (n - 1)] + (2 - m)(2n - mn) - m[2mn - (n - 3)] + (2m - 1) = 0$$
Solving the above equation, we can obtain that $n = \dfrac{2}{5 - 5m}$. Wait, so what if $m = 1$? Then $(P_m) \colon y - z + 1 = 0$ and $M = (1; 1; 2)$. Huh, nothing weird to see here.
Anyhow, for $n = \dfrac{2}{5 - 5m}$, $M = \left(\dfrac{3}{5}; \dfrac{2m - 4}{5m - 5}; \dfrac{11m - 13}{5m - 5}\right)$. Let $M (a, b, c)$ be the coordinates of the centre of the fixed circle, then $\left(a - \dfrac{3}{5}\right)^2 + \left(b - \dfrac{2m - 4}{5m - 5}\right)^2 + \left(c - \dfrac{11m - 13}{5m - 5}\right)^2$ is the measurement of the radius of the circle squared and is a constant. Let's call it $r^2$.
We have that $$\begin{aligned} r^2 &= \left(a - \dfrac{3}{5}\right)^2 + \left(b - \dfrac{2m - 4}{5m - 5}\right)^2 + \left(c - \dfrac{11m - 13}{5m - 5}\right)^2\\ &= (a^2 + b^2 + c^2) - \dfrac{2[(3a + 2b + 11c)m - (3a + 4b + 13c)]}{5m - 5} + \dfrac{2(67m^2 - 160m + 97)}{(5m - 5)^2} \end{aligned}$$
Let's break down each part. First of all, $\dfrac{2[(3a + 2b + 11c)m - (3a + 4b + 13c)]}{5m - 5} = \dfrac{2}{5} - \dfrac{4b + 4c}{5m - 5}$. Additionally, $$\dfrac{2(67m^2 - 160m + 97)}{(5m - 5)^2} = \dfrac{134}{25} - \dfrac{52}{5(5m - 5)} + \dfrac{8}{(5m - 5)^2}$$
That means $$r^2 = \dfrac{124}{25} + (a^2 + b^2 + c^2) + \dfrac{4(5b + 5c - 13)}{5(5m - 5)} + \dfrac{8}{(5m - 5)^2}$$
Since $r^2$ is a constant, that means all such terms involving $m$ should be a constant also, that means $5b + 5c - 13 = 8 = 0$...? Something's wrong here.
That's where it ends for now. And even then, this attempt was accompanied by the help of a laptop. So, you know, it isn't really efficient.
To end it off, I'm asking for a solution which doesn't require much clunky calculation. As always, thanks for reading, (and even more if you could help~) By the way, the options were $\left(1; \dfrac{1}{2}; \dfrac{5}{2}\right); \left(1; -\dfrac{1}{2}; \dfrac{5}{2}\right); \left(1; \dfrac{1}{2}; \dfrac{3}{2}\right); \left(1; -\dfrac{1}{2}; \dfrac{3}{2}\right)$.
Plane equation is $(m-1) x + (2-m) y - m z + (2m-1) = 0$
Point is $A = (1, 0, 3)$
The line perpendicular to plane and passing through $A$ is
$P(t) = A + t ( m-1, 2 - m , -m )$
Plug this into the plane equation you get
$ t = 2 / ( 3m^2 - 6 m + 5 )$
Plug this value of $t$ into equation of line, you get the projection point
$Q(m) = (1, 0, 3) + \dfrac{2}{3m^2 - 6m + 5} ( m-1, 2 - m , -m)\\ = \dfrac{1}{3m^2 - 6 m + 5 } ( 3m^2 - 4 m +3, 4 - 2 m , 9 m^2 - 20 m + 15 )$
Let's check whether Q(m) lies on a single plane. If it does, then we can find constants $a,b,c$ such that
$a x + b y + c z + 1 = 0$
Plug in the $x,y,z$ coordinates of $Q(m)$ into the equation of the plane, you get a quadratic polynomial in $m$,
$a (3 m^2 - 5m + 3) + b (4 - 2m) + c (9 m^2 - 20 m + 15) + (3 m^2 - 6m + 5) = 0 $
Equating the coefficient of $m^2 ,m, 1$ on the left hand side to $0$, we get the linear system
$ 3 a + 9 c = -3\\ - 5 a - 2 b - 20 c = 6\\ 3 a + 4 b + 15 c = -5\\ $
The solution of the linear system is $ a= -0.4 , b = -0.2 , c = -0.2 $
So indeed $Q(m)$ lies on the plane
$-0.4 x - 0.2 y -0.2 z + 1 = 0$
or equivalently
$ 2 x + y + z - 5 = 0 $
A point on this plane is $(0,0,5)$
No we want to project $Q(m)$ on the axes of a reference frame $(O'x'y'z')$ attached to this plane, with $O' = (0,0,5)$ and the $z'$ normal to the plane along the unit vector $\hat{n}=(2, 1, 1)/\sqrt{6} $. The $x',y'$ axes are taken perpendicular to $\vec{n} $ . For example, we can choose
$\hat{x} = (1, -2, 0) / \sqrt{5} $
and it would follow that
$ \hat{y} = \hat{n} \times \hat{x} = (2, 1, -5) /\sqrt{30} $
Now our frame $O'x'y'z'$ is related to the world frame by
$ p = (0,0,5) + R p' $
where $p$ is the coordinate vector in $Oxyz$ and $p' $ is the coordinate vector in $O'x'y'z'$ , and
$R = \begin{bmatrix} \dfrac{1}{\sqrt{5}} && \dfrac{2}{\sqrt{30}} && \dfrac{2}{\sqrt{6} }\\ -\dfrac{2}{\sqrt{5}} && \dfrac{1}{\sqrt{30}} && \dfrac{1}{\sqrt{6}} \\ 0 && -\dfrac{5}{\sqrt{30}} && \dfrac{1}{\sqrt{6}} \end{bmatrix} $
Hence the coordinates of $Q(m)$ in the plane is given by the vector
$Q'(m) = R^T (Q(m) - (0,0, 5) )$
We have
$ Q(m) - (0, 0, 5) = \dfrac{1}{3m^2 - 6 m + 5 } ( 3m^2 - 4m + 3, 4 - 2m , - 6 m^2 + 10 m -10 )\\ =\dfrac{1}{3m^2 - 6 m + 5 } ( 3m^2 - 4m +3, 4 - 2 m , 9 m^2 - 20 m + 15 )$
Hence,
$R^T (Q(m) - (0,0,5) ) =\left(\dfrac{1}{\sqrt{5}}\right) \dfrac{1}{3 m^2 - 6m + 5} ( 3 m^2 - 5, \sqrt{6} ( 6 m^2 - 10 m + 10 ) , 0 ) $
Now, $ 3m^2 - 6m + 5 = 3 (m - 1)^2 + 2 = 2 ( \dfrac{3}{2} (m-1)^2 + 1 ) $
Let $t = \sqrt{\dfrac{3}{2}}(m-1)$, then after simplification, we get
$(x',y') = ( \dfrac{1}{\sqrt{5}} \left( 1 + \dfrac{ \sqrt{6} t - 2 }{ t^2 + 1} \right), \dfrac{1}{\sqrt{5}} \left( 2 \sqrt{6} + \dfrac{2 t + \sqrt{6} } {t^2 + 1 } \right) ) $
Define $\cos(\theta) = \dfrac{1}{\sqrt{t^2 + 1}} , \sin(\theta) = \dfrac{t}{ \sqrt{t^2 + 1} } $
Then,
$(x',y') = ( \dfrac{1}{\sqrt{5}} ( 1 + \sqrt{6} \sin(\theta) \cos(\theta) - 2 \cos^2(\theta) ), \dfrac{1}{\sqrt{5}} ( 2 \sqrt{6} + 2 \sin(\theta)\cos(\theta) + \sqrt{6} \cos^2 (\theta) ) ) \\ = ( \dfrac{1}{\sqrt{5}} \left( \dfrac{ \sqrt{6}}{2} \sin(2 \theta) - \cos(2 \theta)\right) , \dfrac{1}{\sqrt{5}}\left(\dfrac{ 5 \sqrt{6}}{2} + \sin(2 \theta) + \dfrac{\sqrt{6}}{2} \cos( 2 \theta) \right)) $
Define $\phi$ such that $\cos \phi = \sqrt{\dfrac{2}{5}} , \sin \phi = \sqrt{\dfrac{3}{5}} $ then
$(x',y') =\left(\dfrac{1}{\sqrt{5}} \right) ( 0, \dfrac{5 \sqrt{6}}{2} ) + \dfrac{1}{\sqrt{5} } \sqrt{\dfrac{ 5}{2} } ( - \cos( 2 \theta + \phi ) , \sin(2 \theta + \phi) ) )$
which is an equation of a circle with center $C(x',y') = (0, \dfrac{\sqrt{30}}{2} )$, and radius $\dfrac{1}{\sqrt{2}}$ .
Finally, we need to convert the coordinates of the center from the $O'x'y'z'$ frame to the world coordinate system $Oxyz$, using
$ p = (0, 0, 5) + R p' $
This gives, $ C(x,y,z) = (0, 0, 5) + \dfrac{\sqrt{30}}{2} \hat{y} \\ = (0, 0, 5)+ \dfrac{\sqrt{30}}{2} \left(\dfrac{1}{\sqrt{30}}\right) ( 2 , 1, -5 ) \\ = (1, \dfrac{1}{2}, \dfrac{5}{2} ) $