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The equations of $2$ double cones having their vertex at the origin $(0,0,0)$ are given by:

$(ax+by+cz)^2=(x^2+y^2+z^2)\cos^2(\theta_1) \hspace{25pt} (1)$

($\theta_1=$semi-apical angle, and $a,b,c$ are the direction cosine of the axis of the cone)

$(a'x+b'y+c'z)^2=(x^2+y^2+z^2) \cos^2(\theta_2) \hspace{25pt} (2)$

($\theta_2=$ semi-apical angle, and $a',b',c'$ are the direction cosine of the axis of the cone)

Suppose, it's given that the 2 cones intersect. So, it is obvious that their intersection will yield a pair of straight lines (Also at a specific case only a line, when just touching each other).

What will be the equation of the lines (or line)?

Hosam Hajeer
  • 21,978
Niladri Das
  • 11

1 Answers1

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The two given cones share the same vertex which is the origin. The axis of the first is the unit vector $u_1 = (a,b,c)$ while the second's axis is the unit vector $u_2 = (a',b',c')$. The semi-apical angle of the first is $\theta_1$ and the second's is $\theta_2$.

The two lines of the intersection, are those vectors $r = (x,y,z)$ that make an angle of $\theta_1$ with $u_1$ and $ \theta_2$ with $u_2$.

Imposing $r$ to have a unit length results in the following three equations

$ a x + b y + c z = \cos \theta_1 $

$ a' x + b' y + c' z = \cos \theta_2 $

$ x^2 + y^2 + z^2 = 1 $

The solution of this system is relatively easy because the first two equations are linear in $x,y,z$. Their solution is the line

$ L(\lambda) = V_0 + \lambda V_1 $

where $ V_1 = (a,b,c) \times (a',b',c') $ is the direction vector, and

$ V_0 $ is any point that solves both equations. Such a point can be found by setting $z = 0 $ and solving the remaing $2 \times 2$ system. Doing so, gives

$ x_1 = \dfrac{ b' \cos \theta_1 - b \cos \theta_2 }{ a b' - a' b } $

$ y_1 = \dfrac{ a \cos \theta_2 - a' \cos \theta_1 }{a b' - a' b } $

So that

$ V_0 = (x_1, y_1, 0) $

Now plugging $L(\lambda)$ into the third equation, gives

$ (V_0 + \lambda V_1) \cdot (V_0 + \lambda V_1) = 1 $

which is

$ \lambda^2 \ (V_1 \cdot V_1) + 2 \lambda (V_0 \cdot V_1) + (V_0 \cdot V_0 - 1) = 0 $

This will have two possible solutions, which will give the two unit vectors $r$.

As an explicit example, suppose the unit vectors $(a,b,c)$ and $(a',b',c') $ are given by

$ (a,b,c) = \dfrac{1}{3}(1,2,2) $

$ (a',b',c') = \dfrac{1}{9}(1, -4, 8) $

And let

$ \theta_1 = \dfrac{\pi}{3} $

$\theta_2 = \dfrac{\pi}{4}$

Applying the above formulas,

$V_1 = (a,b,c) \times (a',b',c') = \dfrac{1}{9} (8, -2, -2) $

And since this a direction vector, we can scale it, so we'll take

$ V_1 = (4, -1, -1) $

As $V_0$, we have

$x_1 = 1 +\dfrac{3}{2} \sqrt{2} $

$ y_1 = -\dfrac{3}{ 4} \sqrt{2} + \dfrac{1}{4} $

so that $V_0 = (x_1, y_1, 0) $

Using these two vector into the third (quadratic) equation, gives two roots.

$ \lambda_{1,2} = -0.89167 , -0.58565 $

And these two values of $\lambda$ generate the following unit vectors solutions to the problem.

$ u_1 = \begin{pmatrix} -0.44537 \\ 0.081013 \\ 0.891673 \end{pmatrix} \hspace{25pt} u_2 = \begin{pmatrix} 0.778705 \\ -0.22501 \\ 0.585654 \end{pmatrix} $

Hosam Hajeer
  • 21,978