Vector equations with trigonometry

The question states that the tip is at $(0, 5, 0)$, and that the base perimeter fulfills $x^2 + z^2 = 1$.

(Technically speaking, $x^2 + z^2 = 1$ is really a right circular cylinder of infinite length and radius $1$, axis on the $y$ axis. Here, they obviously mean that the cone base is a circle of radius $1$ on the $x z$ plane, centered at origin.)

Base perimeter is a circle of radius $1$ centered at origin $(0, 0, 0)$.

If you have a line that passes through $(0, 5, 0)$ in direction $(0, 1, 0)$, it will pass outwards from the tip. It has already passed through origin, $(0, 0, 0)$. However, origin is at the center of the base; it is not at the perimeter of the base. (Perimeter is the outer boundary.)

Any line that passes through the tip at $(0, 5, 0)$ and a point on the perimeter of the base, say $(1, 0, 0)$, $(0, 1, 0)$, $(-1, 0, 0)$, or $(0, -1, 0)$, will fulfill the rule.

We can parametrise the points on the perimeter using angle parameter $theta$, as $(costheta, 0, sintheta)$. (If $theta = 0$, the point is $(1, 0, 0)$; if $theta = 90°$, the point is $(0, 1, 0)$; if $theta = 180°$, the point is $(-1, 0, 0)$, and so on.)

Any line can be parametrised using a real parameter, say $lambda$, and two points the line passes through, say $vec{a}$ and $vec{b}$: $$ell = (1 - lambda) vec{a} + lambda vec{b} = vec{a} + lambda (vec{b} - vec{a})$$ (This is also the equation for linear interpolation.) When $lambda = 0$, $ell = vec{a}$. When $lambda = 1$, $ell = vec{b}$.

Here, $vec{a} = (0, 5, 0)$ (the tip), and the point on the perimeter $vec{b} = (costheta, 0, sintheta)$, so the equation of the line is $$ell = (1 - lambda)left[begin{matrix}0\5\0end{matrix}right] + lambdaleft[begin{matrix}costheta\0\sinthetaend{matrix}right] = left[begin{matrix}0\5\0end{matrix}right] + lambdaleft[begin{matrix}costheta\-5\sinthetaend{matrix}right]$$