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 $ ?

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 $
