I recently had this question on one of my math tests involving vectors:
Find the exact distance between a point on the line l : $\frac{x+2}{2} = \frac{4-z}{6}, y=5$ and the plane containing the line $(x-5, y+3, z+2) = \mu(1,0,3)$.
I found the question to be very ambiguous since there are an infinite amount of planes containing the line $(x-5, y+3, z+2) = \mu(1,0,3)$. (therefore the must be an infinite amount of planes containing a single line that will intersect the line l and in only one case will they not intersect) This means there is no exact distance between a point on the line l and the plane containing the line $(x-5, y+3, z+2) = \mu(1,0,3)$.
Does anyone have a valid solution for this problem, have I overlooked something?
The first line lies in the $y = 5$ plane; the second lies in the $y = -3$ plane. But I don't think that makes any difference.
Your argument is pretty solid. For any point $Q$ on the first line, there is SOME plane $P$ containing both the second line and $Q$, and for $P$, the distance is zero. But for any point $Q$ on the line, there's also some plane $P'$ containing the second line and \emph{not} containing $Q$ (this actually requires a little proof, and the y-values argument helps), and in this case, the distance is positive.
It's a horribly worded question. I cannot imagine what the questioner was hoping to get as an answer.