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I’m not sure how to proceed with the following question. Could anyone help me? Thanks.
Write down a (vector) equation for the right-angled cone centered around the line $$x = y = z$$ in three-dimensional space by finding an equation satisfied by all vectors (x, y, z) which make an angle of $\pi/4$ radians with the vector $$(1,1,1)$$ which points along that line.
:/ Any help would be greatly appreciated. (:
With the vertex at the origin, then the position vectors $\vec{r}$ of all points on the cone satisfy $$ \vec{r}.\hat{n}=|\vec{r}|\cos\pi/4 $$ where $\hat{n}$ is the unit vector in the axis direction $(1,1,1) $ and $\cos\pi/4$ is well known.
For $(x,y,z)$ to make an angle of $\pi /4$ with $(1,1,1)$ we need to have
$$\cos (\pi /4) = \frac {x+y+z}{\sqrt 3 \sqrt {x^2+y^2+z^2} }$$
Upon simplification we get $$ x^2 + y^2 + z^2 = 4(xy+yz+xz)$$
Which is the equation of the desired cone in Cartesian coordinate.
HINT
The equation of a right circular cone with vertex at the origin, axis parallel to the vector $d$ and aperture $2\theta$ is given by
$$F(u) = (u \cdot d)^2 - (d \cdot d) (u \cdot u) \cos^2 \theta$$
or
$$F(u) = u \cdot d - |d| |u| \cos \theta$$
where $u=(x,y,z)$.
HINT
Beginning with $$ z = r = \sqrt{x^2+y^2} $$
can you rotate/shift cone axis along diagonal of unit cube?
