$OABC$ is a parallelogram with $O$ at the origin and $a,b,c$ are the position vectors of the points $A,B, and$ $C$.

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$OABC$ is a parallelogram with $O$ at the origin and $a,b,c$ are the position vectors of the points $A,B, and$ $C$. $P$ is the midpoint of $BC$ and $Q$ is the point on $OB$ such that $OQ:QB$ is 2:1.
Prove $APQ$ is a straight line



I keep seeming to come round in circles in trying to prove this and it is just seeming unnecessarily messy. Any help would be appreciated.






vectors







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  • What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
    – Lord Shark the Unknown
    Sep 6 at 18:55










  • added it in. sorry i missed a bit off
    – H.Linkhorn
    Sep 6 at 18:57










  • $b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
    – Lord Shark the Unknown
    Sep 6 at 19:02










  • But if I do that there isn't a common scalar for both c and b terms
    – H.Linkhorn
    Sep 6 at 19:10














up vote
1
down vote

favorite












$OABC$ is a parallelogram with $O$ at the origin and $a,b,c$ are the position vectors of the points $A,B, and$ $C$. $P$ is the midpoint of $BC$ and $Q$ is the point on $OB$ such that $OQ:QB$ is 2:1.
Prove $APQ$ is a straight line



I keep seeming to come round in circles in trying to prove this and it is just seeming unnecessarily messy. Any help would be appreciated.






vectors







share|cite|improve this question























  • What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
    – Lord Shark the Unknown
    Sep 6 at 18:55










  • added it in. sorry i missed a bit off
    – H.Linkhorn
    Sep 6 at 18:57










  • $b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
    – Lord Shark the Unknown
    Sep 6 at 19:02










  • But if I do that there isn't a common scalar for both c and b terms
    – H.Linkhorn
    Sep 6 at 19:10












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$OABC$ is a parallelogram with $O$ at the origin and $a,b,c$ are the position vectors of the points $A,B, and$ $C$. $P$ is the midpoint of $BC$ and $Q$ is the point on $OB$ such that $OQ:QB$ is 2:1.
Prove $APQ$ is a straight line



I keep seeming to come round in circles in trying to prove this and it is just seeming unnecessarily messy. Any help would be appreciated.






vectors







share|cite|improve this question















$OABC$ is a parallelogram with $O$ at the origin and $a,b,c$ are the position vectors of the points $A,B, and$ $C$. $P$ is the midpoint of $BC$ and $Q$ is the point on $OB$ such that $OQ:QB$ is 2:1.
Prove $APQ$ is a straight line



I keep seeming to come round in circles in trying to prove this and it is just seeming unnecessarily messy. Any help would be appreciated.






vectors




vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 6 at 18:57

























asked Sep 6 at 18:53









H.Linkhorn

358




358











  • What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
    – Lord Shark the Unknown
    Sep 6 at 18:55










  • added it in. sorry i missed a bit off
    – H.Linkhorn
    Sep 6 at 18:57










  • $b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
    – Lord Shark the Unknown
    Sep 6 at 19:02










  • But if I do that there isn't a common scalar for both c and b terms
    – H.Linkhorn
    Sep 6 at 19:10
















  • What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
    – Lord Shark the Unknown
    Sep 6 at 18:55










  • added it in. sorry i missed a bit off
    – H.Linkhorn
    Sep 6 at 18:57










  • $b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
    – Lord Shark the Unknown
    Sep 6 at 19:02










  • But if I do that there isn't a common scalar for both c and b terms
    – H.Linkhorn
    Sep 6 at 19:10















What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
– Lord Shark the Unknown
Sep 6 at 18:55




What's the problem? You define a bunch of points, but don't say what it is you need to prove about them.
– Lord Shark the Unknown
Sep 6 at 18:55












added it in. sorry i missed a bit off
– H.Linkhorn
Sep 6 at 18:57




added it in. sorry i missed a bit off
– H.Linkhorn
Sep 6 at 18:57












$b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
– Lord Shark the Unknown
Sep 6 at 19:02




$b=a+c$, $p=frac12(b+c)$ and $q=frac13b$. You need to show that $p-a$ is a scalar multiple of $q-a$.
– Lord Shark the Unknown
Sep 6 at 19:02












But if I do that there isn't a common scalar for both c and b terms
– H.Linkhorn
Sep 6 at 19:10




But if I do that there isn't a common scalar for both c and b terms
– H.Linkhorn
Sep 6 at 19:10










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If you mark midpoint of $AC$ with $R$ you see that $R$ is also a midpoint for $OB$ since $OABC$ is a paralellogram.



So $Q$ is a gravity center for $ABC$ since $BQ:QR=2:1$, so $Q$ is also on $AP$.






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    If you mark midpoint of $AC$ with $R$ you see that $R$ is also a midpoint for $OB$ since $OABC$ is a paralellogram.



    So $Q$ is a gravity center for $ABC$ since $BQ:QR=2:1$, so $Q$ is also on $AP$.






    share|cite|improve this answer
























      up vote
      0
      down vote













      If you mark midpoint of $AC$ with $R$ you see that $R$ is also a midpoint for $OB$ since $OABC$ is a paralellogram.



      So $Q$ is a gravity center for $ABC$ since $BQ:QR=2:1$, so $Q$ is also on $AP$.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        If you mark midpoint of $AC$ with $R$ you see that $R$ is also a midpoint for $OB$ since $OABC$ is a paralellogram.



        So $Q$ is a gravity center for $ABC$ since $BQ:QR=2:1$, so $Q$ is also on $AP$.






        share|cite|improve this answer












        If you mark midpoint of $AC$ with $R$ you see that $R$ is also a midpoint for $OB$ since $OABC$ is a paralellogram.



        So $Q$ is a gravity center for $ABC$ since $BQ:QR=2:1$, so $Q$ is also on $AP$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 19:19









        greedoid

        28.7k93878




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