# Dot Product Cone

A circular cone has its vertex at the origin and its axis in the direction of the unit vector $$\hat{a}$$. The half-angle at the vertex is $$\alpha$$. Show that the position vector $$r$$ of a general point on its surface satisfies the equation

$$\hat{a} \cdot r = | r| \cos(\alpha)$$

Obtain the cartesian equation when $$\hat{a} = (2/7, −3/7, −6/7)$$ and $$\alpha = 60^\circ$$.

I have no idea how to do the first part to prove the formula. For the second part I know that the direction vector is therefore $$2i - 3j - 6k$$.

Since the cone passes through the origin $$(0,0,0)$$, could I write the cartesian equation as $$x2 = y-3 = z-6$$ ?

• Oh, because it says general point on the surface, does that mean I want to be finding the cartesian equation of the plane and not the line ?
– user1071088
Jul 1, 2022 at 11:11
• My comments were in error, as they applied only to first part. Deleting. Jul 1, 2022 at 14:58

The vector from the origin to a point $$r$$ on the surface of the cone makes an angle $$\alpha$$ with the axis vector $$\hat{a}$$, therefore, using dot product,

$$\cos(\alpha) = \dfrac{ r \cdot \hat{a} } { \| r \| \| \hat{a}\| }$$

Since $$\hat{a}$$ is a unit vector , then $$\| \hat{a} \| = 1$$ , and the above equation becomes

$$r \cdot \hat{a} = \| r \| \cos(\alpha)$$

To obtain the cartesian equation, square both sides of the above equation, to obtain

$$(r \cdot \hat{ a} ) ( r \cdot \hat{a} ) = \| r \|^2 \cos^2(\alpha)$$

Now using linear algebra notation for vector and the transpose operation, we have

$$r \cdot \hat{a} = r^T \hat{a} = \hat{a}^T r$$

Therefore, we now have

$$r^T \hat{a} \hat{a}^T r = r^T r \big( \cos^2(\alpha) \big)$$

Taking $$r^T$$ and $$r$$ as common factors on the left and right, the equation becomes

$$r^T \big( \cos^2(\alpha) I - \hat{a} \hat{a}^T \big) r = 0$$

where $$I$$ is the identity matrix. And this is of the form

$$r^T Q r = 0$$ with $$Q =\cos^2(\alpha) I - \hat{a} \hat{a}^T$$

Now if $$\hat{a} = ( 2/7, -3/7, -6/7 )$$ then $$\hat{a}$$ is a unit vector because $$2^2 + 3^2 + 6^2 = 7^2$$, and since $$\alpha = 60^\circ$$, then $$\cos(\alpha) = \dfrac{1}{2}$$

Therefore,

$$Q = \dfrac{1}{4} I - \dfrac{1}{49} \begin{bmatrix} 4 && - 6 && -12 \\ -6 && 9 && 18 \\ -12 && 18 && 36 \end{bmatrix} = \dfrac{1}{196} \begin{bmatrix} 33&& 24 && 48 \\ 24 && 13 && - 72 \\ 48 && -72 && -95 \end{bmatrix}$$

Therefore, since $$r = [x, y, z]^T$$ , then the cartesian equation of the cone is

$$33 x^2 + 13 y^2 - 95 z^2 + 48 xy + 96 xz - 144 yz = 0$$