Let $O,A$ be two points in the plane with $|\vec{OA}|=3$. Which line is formed by the points $M$ of the plane, such that $\vec{OM}(\vec{OM}-2\vec{OA})=7$ ?

My attempt..

Suppose $\vec{OM}=\vec{x}$ and $\vec{OA}=\vec{y}$. Then the above equation can be rewritten as $C: \vec{x}^2-2\vec{x}\vec{y}-7=0$. The discriminant is $64$, thus $\vec{x}=\vec{y}\pm 4\implies \vec{OM}=\vec{OA}\pm 4$. So for every value of $\vec{OA}$ there are two values of $\vec{OM}$..

Here's where i'm stuck. I know that $C$ is a parabola, so the points $M$ must lie in a parabola?? I want to understand what i have to do here and in general i want to know if what i did makes sense.. What have i accomplished really? I'm confused. Please help. Thanks in advance.

  • $\begingroup$ Just to make sure, according to the problem, $OA=3$, correct? Therefore, couldn't we just say $OM=3 \pm 4$ and thus, excluding negative solutions $OM=7$? Once we have that, we can say that the set of possible points for $M$ is a circle of radius $7$ around $O$. $\endgroup$ Jun 23, 2016 at 12:11
  • $\begingroup$ The modulus of $OA$ is 3, not $\vec{OA}$.. $\endgroup$
    – KeyC0de
    Jun 23, 2016 at 12:13
  • $\begingroup$ Oh, OK. Does this mean that the $\cdot$ in your equation is a dot product? $\endgroup$ Jun 23, 2016 at 12:14
  • $\begingroup$ I'm still a bit confused as to what you're doing. If this isn't a dot product than what is it? Also, how can you add or subtract a vector by a scalar like you are doing now? It seems like you used the quadratic formula with $-2\vec y$ as your $b$ coefficient, but this makes no sense because your're mixing vectors up with scalars. $\endgroup$ Jun 23, 2016 at 12:36
  • 1
    $\begingroup$ You say $\vec y \pm 4$. That's the part that doesn't make sense: How do you add a vector and a scalar? Also, your new answer makes much more sense, in my opinion. You should post that as an answer to your own question (yes, you can answer your own question). Then, after two days, you can select that as the accepted answer. That way, this won't come up in the unanswered queue. $\endgroup$ Jun 23, 2016 at 13:09

1 Answer 1


I found the answer. Here goes..

By letting $A$ be the point of reference we have:

$\vec{OM}(\vec{OM}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}-\vec{AO}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}+\vec{AO})=7 \implies |\vec{AM}|^2 -|\vec{AO}|^2=7 \implies |\vec{AM}|^2=7+9=16 \implies |\vec{AM}|=4$

Therefore we notice that the points $M$ are located $4$ units around $A$, which means they are in a circle with center $A$ and radius $4$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .