Question :- Prove that if a plane cut the cone $$ax^2+by^2+cz^2=0$$ in perpendicular generators, it touches the cone $$\frac{x^2}{b+c}+\frac{y^2}{a+c}+\frac{z^2}{a+b}=0$$
In solution :- Consider a plane $$ux+vy+wz=0$$ Now , consider a section of cone and plane is $$\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$$.
Since this lies in cone and plane so must satisfy these equation.Therefore,
$$ul+vm+wn=0, al^2+bm^2+cn^2=0—–(3)$$. These two line are perpendicular if $$u^2(b+c)+v^2(a+c)+w^2(a+b) = 0 —-(4)$$
Here, in this step i got somewhat confused :- $$$$
1. When we satisfy the cone and plane equation by replacing with l,m,n in (3), this should just be an equation, then why we are treating as line and work on their perpendicularity.?$$$$
2. From where it hold that if two line are perpendicular $$ul+vm+wn=0, al^2+bm^2+cn^2=0—–(3)$$ then $$u^2(b+c)+v^2(a+c)+w^2(a+b) = 0 —-(4)$$ should be true.?
I never studied such concept. Please clarify it will be a great help.
Best Answer
That is the equation of a line (passing through the origin), not a section. Note that $x, y, z$ are the variables here. $l, m, n$ are fixed parameters defining which line the equation is for. Further, the values of the parameters $l, m, n$ for a given line are only unique up to a common multiple. I.e., if you multiply all three by the same value, the three new values will still determine the same line.
The same is true of the plane and cone equations: In $$ux + vy + wz = 0$$ $(x,y,z)$ are the coordinates of points on the plane, while $u, v, w$ are fixed parameters determining which plane it is, and are only determined up to a multiple. In $$ax^2 + by^2 + cz^2 = 0$$ $(x,y,z)$ are the coordinates of points on the elliptical cone (cross-sections are ellipses), while $a, b, c$ are parameters defining the cone, and again, are only unique up to a multiple.
All these multiples can be problematic, so I am going to make some additional assumptions: $$u^2 + v^2 + w^2 = 1\\a^2 + b^2 + c^2 = 1\\l^2 + m^2 + n^2 = 1$$ As long as the parameters are not all $0$ (which would mean you don't actually have a cone, plane, or line) you can always multiply them by a value that makes these equations true. With these restrictions, the only multiplier possible is $-1$. To remove even this possibility, I will also require that $n > 0$.
Now a section of the cone by the plane is not one line, but two lines. That is, there is a line $\frac xl = \frac ym = \frac zn$, and a second line $\frac x{l'} = \frac y{m'} = \frac z{n'}$ which are both contained in the intersection of the plane and the cone. Since $(x,y,z) = (l, m, n)$ is a point on the first line and $(x,y,z) = (l', m', n')$ is a point on the second line, these two points must themselves be in both the plane and the cone:
$$ul + vm + wn = 0\\al^2+bm^2 + cn^2 = 0$$ $$ul' + vm' + wn' = 0\\al'^2+bm'^2 + cn'^2 = 0$$
We solve for $l, m, n$ first. There are three equations in these three unknowns: $$ul + vm + wn = 0\\al^2+bm^2 + cn^2 = 0\\l^2+m^2 + n^2 = 1$$
If we solve the first for $l$ and substitute in the other two: $$(bu^2 + av^2)m^2 + 2avwmn + (cu^2 + aw^2)n^2 = 0\\(u^2 + v^2)m^2 + 2vwmn + (u^2 + w^2)n^2 = u^2$$ Letting $r = \frac mn$, we can rewrite the equations
$$(bu^2 + av^2)r^2 + 2avwr + (cu^2 + aw^2) = 0\\(u^2 + v^2)r^2 + 2vwr + (u^2 + w^2) = u^2n^2$$ The first equation is a quadratic in $r$ with solutions $$r = \frac{-avw \pm u\sqrt{-(bcu^2 +abw^2 + acv^2)}}{bu^2 + av^2}$$
Once $r$ is determined, we can find the values of $l, m, n$:
$$n = \sqrt{\frac{(u^2 + v^2)r^2 + 2vwr + (u^2 + w^2)}{u^2}}\\ m = rn\\ l = -\frac {vm + wn}u$$
Now there are two choices for $r$. One gives $l,m,n$, the other gives the other line $l', m', n'$
The two lines will be perpendicular if $$ll' + mm' + nn' = 0$$ (Since $l^2 + m^2 + n^2 = 1$ and $l'^2 + m'^2 + n'^2 = 1$, you can show that $ll' + mm' + nn'$ is the cosine of the angle between the two lines. When they are perpendicular, that angle is $\pi/2$, and the cosine is $0$.)
If you plug in the expressions for $l, m, n, l', m', n'$ above and do a lot of simplification, the claim is that the result will be $$u^2(b+c)+v^2(a+c)+w^2(a+b)=0$$
This a condition that the plane $ux + vy + wz = 0$ and cone $ax^2 + by^2 + cz^2 = 0$ must satisfy if their intersection is going to consist of perpendicular generators, instead of generators meeting in some other angle.
(The fact that it requires so much tedious algebra means that this was not the best way to produce this result. You will find that pretty much every difficult calculation you encounter in mathematics can be reduced to something relatively trivial, if only you knew the right way to look at it. Alas, the right way to look at it usually only becomes apparent months or years after you've performed the calculation.)