A Planar Cut through an Oblique Cone

Vectors connecting the apex to points on the base along the lateral curved surface of the cone are given by

$v(t) = (5 \cos t, 5 \sin t - 5, -20)$

All the other points on the surface of the cone are scalar multiples of these vectors plus an offset of the apex coordinates.

We can think of this cone as a linear transformation of another arbitrary standard right circular cone.

Note first that $v(t)$ can be written as

$v(t) = \cos t (5 , 0, 0) + \sin t (0, 5, 0 ) + (0, -5, -20) = v_1 \cos t + v_2 \sin t + v_0$

An arbitrary standard right circular cone with apex at the origin, and a circular base centered at $(0, 0, -h)$ , and radius of base $r$ has the following form of its lateral vectors

$u(t) = \cos t (r, 0, 0) + \sin t (0, r, 0) + (0, 0, -h) = u_1 \cos t + u_2 \sin t + u_0$

So we can write the vectors $(5, 0, 0), (0, 5, 0), (0, -5, -20)$ (which are $v_1, v_2, v_0$ ) as a linear transformation (matrix multiplication) of

the vectors $(r, 0, 0), (0, r, 0),$ and $(0, 0, -h)$ , ( which are $u_1, u_2, u_0$ )

i.e. we set the matrix equation,

$V = A U$

where $V = \begin{bmatrix} v_1 && v_2 && v_0 \end{bmatrix}$ and $U =\begin{bmatrix} u_1 && u_2 && u_0 \end{bmatrix}$

The solution for A is simply $A = V U^{-1}$

For computation, we can take $h = 1, r = 1$ . It follows that points on our oblique cone are given by $r = r_0 + A r'$ , where $r_0$ is the apex of the oblique cone, and $r'$ is the position vector of a point on the right circular cone. Now, we know that the algebraic equation of points on the right circular cone are given by $r'^T Q' r' = 0$ , where for the simple choice that we made of $h=1$ , and $r = 1$ , we have,

$Q' = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix}$

Substituting for $r'$ , we obtain,

${( A^{-1} (r - r_0) )}^T Q' (A^{-1} (r - r_0) ) = 0$

which simplifies to,

$(r - r_0)^T Q (r - r_0) = 0$ where $Q = A^{-T} Q' A^{-1}$

Now, we are ready to intersect this oblique cone with the plane. The vector equation of the plane is

$r = r_1 + W u$

where $r_1$ is any point on the plane, and $W = [ w_1 , w_2 ]$ is a $3 \times 2$ matrix whose columns are arbitrary unit vectors that together with the normal to the plane form an orthonormal basis for $\mathbb{R}^3$ .

Substituting this vector expression into the equation of the oblique cone, one obtains,

$( r_1 - r_0 + W u )^T Q (r_1 - r_0 + W u) = 0$

Expanding this quadratic form gives,

$u^T (W^T Q W) u + 2 u^T W^T Q (r_1 - r_0) + (r_1 - r_0)^T Q ( r_1 - r_0) = 0$

Take $u_0 = - (W^T Q W)^{-1} W^T Q (r_1 - r_0 )$ , then the above equation becomes,

$(u - u_0)^T (W^T Q W) (u - u_0) = C$

where the constant $C$ is given by $C = - (r_1 - r_0)^T Q (r_1 - r_0) + {u_0}^T (W^T Q W) u_0$

Dividing by $C$ , we obtain,

$(u - u_0)^T Q_1 (u - u_0) = 1$

where $Q_1 = \dfrac{1}{C} (W^T Q W)$ .

The final step is to diagonalize $Q_1$ into $Q_1 = R D R^T$ . The diagonal elements the diagonal matrix $D$ are the eigenvalues of $Q_1$ which are the reciprocal of the square of the semi-minor and semi-major axes lengths of the resulting ellipse of intersection.

Note that depending the orientation of the cutting plane, the intersection can also be parabolic or hyperbolic.