At what times, t, does an airplane (vector defined) intersect with a radar beam (2D plane defined)? A radar beam can be defined as a plane in $3D$ space.  If the beam is moving such that the basis is $[1\text{ }  0 \text{ } \cos(wt)]^T$ and $[1 \text{ } \sin(wt) \text{ } 0]^T$ where $t=time$.  Determine the times when an aircraft that is flying a path defined by $[\cos (wt)\text{ }  \sin(wt)  \text{ } 1]$ is in the radar beam.  
My thoughts:  the airplane will intersect the plane formed by the basis at the origin at certain times, $t$.  Therefore, I should form a matrix, $A$, containing all three vectors and solve for $Ax=0$ where $x$ would be the time, $t$ in this situation.  I'm confused however that the time, $t$, is included in these vectors.  
Is my thinking off base?  Any suggestions would be greatly appreciated.  
 A: Let $a=\cos(\omega t)$ and $b=\sin(\omega t)$. Also, let $r_1=[1,0,a]^T$, $r_2=[1,0,b]^T$ and $p=[a,b,1]$. The aircraft is in the range of the radar when $p\in\operatorname{span}\left\{r_1,r_2\right\}$. i.e. $p=c_1r_1+c_2r_2$, for some $c_1,c_2$. Writing out the equations in full gives
$$
\begin{aligned}
a&=c_1+c_2,\\
b&=bc_2,\\
1&=ac_1.
\end{aligned}
$$
The second and third equations give $c_2=1\ (b\neq0)$ and $c_1=1/a\ (a\neq0)$. We will consider the cases $b=0$ and $a=0$ later. Substituting into the first equation simplifies to $a^2-a-1=0$, which has solutions $a=\dfrac{1\pm\sqrt{5}}{2}$. Now, since $a=\cos(\omega t)$, we must have $\cos(\omega t)=\dfrac{1-\sqrt{5}}{2}$, since $\dfrac{1+\sqrt{5}}{2}>1$. This gives
$$
t=\frac{\cos^{-1}\left(\frac{1-\sqrt{5}}{2}\right)}{\omega}\approx\frac{2\pi k\pm2.3370}{\omega},\ k\in{\mathbb N}.
$$
In the case $a=0$, the equation $1=ac_1$ has no solution.
In the case $b=0$, the equation $b=bc_2$ permits this solution.Now, $b=\sin(\omega t)=0$ implies $t=\dfrac{n\pi}{\omega}, n\in{\mathbb Z}$. Furthermore, we have $a^2+b^2=1$. Since $b=0$, $a=\cos(\omega t)=\cos(n\pi)=(-1)^n$. Thus, this condition also holds. Returning to the system of equations, the second is satisfied with $b=0$, giving $c_2$ arbitrary. The third equation gives $c_1=1/a=(-1)^n$. Substituting this information into the first equation: $(-1)^n=(-1)^n+c_2$ gives $c_2=0$.
The complete set of solutions is
$$
t=\frac{n\pi}{\omega},\ n\in\mathbb{N} \text{ and } t\approx\frac{2\pi k\pm2.3370}{\omega},\ k\in{\mathbb N}.
$$
Here is some MATLAB script to visualise the situation:
clear,clc

omega = 2;

% vectors for basis
r1 = @(t_) [1;0;cos(omega*t_)];
r2 = @(t_) [1;sin(omega*t_);0];
p = @(t_) [cos(omega*t_);sin(omega*t_);1];

% times to display the scenario
nt = 1001;
trng = linspace(0,10,nt);

P = zeros(3,4);
c = 5*[1,1;-1,1;-1,-1;1,-1];
T=delaunay(c);

for i = 1:nt
    t = trng(i);

    v1 = r1(t); v2 = r2(t);
    pv = p(t);

    % get some points on the plane
    for j = 1:4
        P(:,j) = v1*c(j,1)+v2*c(j,2);
    end


    ts = trisurf(T,P(1,:),P(2,:),P(3,:),[1,1,1,1]);
    hold on
    plot3(pv(1),pv(2),pv(3),'b.','MarkerSize',20)
    alpha(ts,0.5);
    axis([-5,5,-5,5,-5,5])
    hold off
    pause(0.05)
end

