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$\vec a,\vec b,\vec c,\vec d$ are four non zero vectors and points $P(\vec a), \; Q(\vec a+\vec b+\vec c), \; R(\vec a-\vec b-\vec c), \; S(\vec d),\; T(\vec a +\vec d)$ are distinct points such that T does not lie in the plane of P,Q and S. Find the number of planes equidistant from all the points $P,Q,R,S,T$.

Any three vectors are coplaner, now $P,Q,R$ are collinear as, $\vec{OR}=\dfrac{2\vec{OP}-\vec{OQ}}{ 2-1}$. So any plane parallel to this line will be equidistant from $P,Q,R$. But I think that points $S,T$ may be aligned in space such that we get no plane which is equidistant from all five points. But answer is given $3$ planes are possible. Can not visualise at this stage. Please help.

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    $\begingroup$ Hint: project everything onto the plane perpendicular to the line which contains the three points. $\endgroup$
    – Polygon
    Commented Mar 21 at 18:52
  • $\begingroup$ Collinear has two letters L. I don't know why either. $\endgroup$
    – Will Jagy
    Commented Mar 21 at 18:59
  • $\begingroup$ As suggested I am considering 3 parallel plane $\perp$ to the line $PQR$ each seperated by some distance; one containing Origin $(0,0)$ , 2nd $\vec S$ and 3rd $\vec T$. What to do now ? Need some more guidance. $\endgroup$
    – Skdmg
    Commented Mar 21 at 19:48
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    $\begingroup$ Project everything onto one single plane, not three different planes. For a visual intuition, you can imagine yourself looking directly in the direction of the line $PQR$ (with an orthogonal camera). Each plane parallel to that line will appear as a line. The three collinear points will just appear as a single point, so the 5 points will appear to be just 3. Thus the problem reduces to counting the number of lines equidistant from three points in 2D space $\endgroup$
    – Polygon
    Commented Mar 21 at 21:56
  • $\begingroup$ Getting your point. Now there are $3$ lines possible equidistant from three points in 2d space, each bisecting one of three altitudes of the triangle formed. Thanks for your help! $\endgroup$
    – Skdmg
    Commented Mar 22 at 3:49

1 Answer 1

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$P,Q,R$ are collinear. Let $d=PQ$, with $R\in d $. Then there are $3$ cases to consider :

  • Let $\mathcal P$ the plane containing $d$ and $S$, $S'$ the orthogonal projection of $S$ on $\mathcal P$, $I$ the midpoint of $SS'$ and $\mathcal P_1$ the plane parallel to $\mathcal P$ passing through $I $;
  • Let $\mathcal Q$ the plane containing $d$ and $T$, $T'$ the orthogonal projection of $T$ on $\mathcal Q$, $J$ the midpoint of $TT'$ and $\mathcal P_2$ the plane parallel to $\mathcal Q$ passing through $J $;
  • Let $e:=ST$ and $K$ the midpoint of the common perpendicular to $d$ and $e$ (see here for example) and let $\mathcal P_3$ the plane passing throug $K$, whose direction is the one spaned by that of $d$ and $e$.

$\mathcal P_1,\mathcal P_2,\mathcal P_3$ are the $3$ planes.

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