# Normal to the plane under the condition describes the cone

The plane $lx+my+nz=0$ moves in such a way that its intersection with the planes $ax+by+cz+d=0$ and $a'x + b'y + c'z+d'=0$ are perpendicular. Show that the normal to the plane through the origin describes in general, a cone of the second degree and find its equation.

My analysis

Here the given plane $lx+my+nz=0$ passes through the origin, so considering a normal dropped from origin is an incorrect term

Where am I going wrong?

Soham

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It's still correct. There are various planes $lx+my+nz=0$ that satisfy the hypothesis. The direction ratios of the normal to the plane are $(l,m,n)$. So they are just trying to say the straight line joining the points $(0,0,0)$ and $(l,m,n)$ describes a cone as the plane varies! Also, you could safely assume $d=d'=0$, because then the two corresponding planes then undergo parallel translation, which doesn't affect orthogonality condition in the hypothesis. – Host-website-on-iPage Jul 14 '12 at 10:34
Apparently there are three conditions if I am to assume the statement is right $\sum al =0$ ; $\sum a'l =0$ and $\sum l^{2} =1$ Also to prove that its a cone, my idea is if I can prove that that the foot of the normal follows a curve which is a 2 degree equation in l,m,n then I can safely conclude that the normal is the generator, origin is the vertex and the surface thus etched is a cone. Please do let me know if I am right at this juncture or not. If indeed, then how to go about finding the foot of the normal? – Soham Jul 14 '12 at 10:53
I think what they're saying is that if $L$ is the line of intersection with first plane and $M$ the line of intersection with the second, then $L$ and $M$ are perpendicular to one another. – Host-website-on-iPage Jul 14 '12 at 10:56
Hmm... now this makes some sense. To be frank, I was still struggling to make sense of your first comment. – Soham Jul 14 '12 at 10:57

The intersection with first plane gives the following line (arrive at this just by linear algebra):

$$\frac x{cm-bn}=\frac y{an-cl}=\frac z{bl-am}$$

Similarly the intersection with the second line is:

$$\frac x{c'm-b'n}=\frac y{a'n-c'l}=\frac z{b'l-a'm}$$

For these lines to be perpendicular the direction ratios' inner product should be 0. i.e., $$(cm-bn)(c'm-b'n)+(an-cl)(a'n-c'l)+(bl-am)(b'l-a'm)=0$$ i.e., $$(bb'+cc')l^2+(cc'+aa')m^2+(aa'+bb')n^2-(ab'+a'b)lm-(bc'+b'c)mn-(ac'+a'c)ln=0$$

Try and show that this is the equation of a cone.

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Oh wow! Yes, indeed!thanks for the effort. This indeed is a cone because its a homogenous equation and there are no individual terms of $l,m,n$ in the equation. Many thanks. This equation shows that for the given condition to satisfy the plane has to move in such a way that $l,m,n$ the directional cosines of the normal (and thus the normal) will trace a cone. Thanks – Soham Jul 14 '12 at 11:03
You're welcome :-) – Host-website-on-iPage Jul 14 '12 at 11:19
Is there a way to contact you? Would like some help on a specific thing – Soham Jul 17 '12 at 7:29
Check mail. If you wish to delete it, you can. Will take the conversation on email – Soham Jul 17 '12 at 15:42