A spaceship is traveling through space at constant speed along a straight line that passes through the points $A = (-3,-8,-6)$ and $B = (-8,-4,-3)$.
The star Gliese $061$, located at $P = (-12,0,0)$ is orbited by an inhabited planet. If the spaceship passes within distance $1/2$ of this star, the inhabitants will detect it. Is the spaceship detected?
I'm not sure how to solve this question as usually I get two vector functions which I can then equate.
consider a similar/simpler problem $A=(-2,-5)$, $B=(9,4)$ and $P=(0,1)$, where $A$,$B$ and $P$ mean the same as in your question and the detection radius is still $\frac{1}{2}$
You can parameterise the path of the spaceship, in terms of the time, $t$, and we say it arrives at time $t=1$, as $\vec{r}=(11t-2,9t-5)$.
Now the distance, $d$, from point $p_1=(p_{1x},p_{1y})$ to $p_2=(p_{2x},p_{2y})$ is $$D=\sqrt{(p_{1x}-p_{2x})^2+(p_{1y}-p_{2y})^2}$$
so the distance from $\vec{r}$ to $p$ is $$D(t)=\sqrt{(11t-2-0)^2+(9t-5-1)^2}=\sqrt{(11t-2)^2+(9t-6)^2}$$
So we can differentiate the expression for $D(t)$, with respect to $t$, $$\frac{d}{dt}D(t)=\frac{202t-76}{\sqrt{(11t-2)^2+(9t-6)^2}}$$
so the closest distance is when $\frac{d}{dt}D(t)=0$, therefore $0=202(t)-76$, therefore $t=\frac{76}{202}=\frac{38}{101}$ and $D_{min}=D(\frac{38}{101})=3.3773$.
Therefore it is greater than the detection distance.
hopefully that helps
(I'll see if i can include a method that doesn't use calculus).