# Vector equations with trigonometry

Mathematics Asked by elyx on February 26, 2021

The question

Part a asks to explain why the vector equation for L1 is the written one. I fail to understand why. Any help? It states that L1 passes through the tip and base perimeter of the cone, but shouldn’t the Direction Vector be (0j, 1i, 0k ) in that case?

Thank you.

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 $$ell = (1 - lambda)left[begin{matrix}0\5\0end{matrix}right] + lambdaleft[begin{matrix}costheta\0\sinthetaend{matrix}right] = left[begin{matrix}0\5\0end{matrix}right] + lambdaleft[begin{matrix}costheta\-5\sinthetaend{matrix}right]$$

Answered by Glärbo on February 26, 2021