Proof related to vectors: collinear points, position vectors












1












$begingroup$


I'm wondering if the following sentence is correct:



"Given 3 collinear points (A,B,C) and the relative distances (m and n), the position vector OC of the central point is mOA+nOB."



enter image description here




EDIT: mAB+nAB=AB --> m+n=1.



Thank you so much for your time.






vectors education







share|cite|improve this question











$endgroup$












  • $begingroup$
    When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
    $endgroup$
    – Chris Culter
    Dec 20 '18 at 9:57










  • $begingroup$
    Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 9:59


















1












$begingroup$


I'm wondering if the following sentence is correct:



"Given 3 collinear points (A,B,C) and the relative distances (m and n), the position vector OC of the central point is mOA+nOB."



enter image description here




EDIT: mAB+nAB=AB --> m+n=1.



Thank you so much for your time.






vectors education







share|cite|improve this question











$endgroup$












  • $begingroup$
    When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
    $endgroup$
    – Chris Culter
    Dec 20 '18 at 9:57










  • $begingroup$
    Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 9:59
















1












1








1





$begingroup$


I'm wondering if the following sentence is correct:



"Given 3 collinear points (A,B,C) and the relative distances (m and n), the position vector OC of the central point is mOA+nOB."



enter image description here




EDIT: mAB+nAB=AB --> m+n=1.



Thank you so much for your time.






vectors education







share|cite|improve this question











$endgroup$




I'm wondering if the following sentence is correct:



"Given 3 collinear points (A,B,C) and the relative distances (m and n), the position vector OC of the central point is mOA+nOB."



enter image description here




EDIT: mAB+nAB=AB --> m+n=1.



Thank you so much for your time.






vectors education




vectors education






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 9:59







Gennaro Arguzzi

















asked Dec 20 '18 at 9:50









Gennaro ArguzziGennaro Arguzzi

361314




361314












  • $begingroup$
    When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
    $endgroup$
    – Chris Culter
    Dec 20 '18 at 9:57










  • $begingroup$
    Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 9:59




















  • $begingroup$
    When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
    $endgroup$
    – Chris Culter
    Dec 20 '18 at 9:57










  • $begingroup$
    Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 9:59


















$begingroup$
When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
$endgroup$
– Chris Culter
Dec 20 '18 at 9:57




$begingroup$
When you say "relative distances", do you mean that $m+n=1$? Either way, make sure that your prediction handles the limiting case where $m=0$ and $n=1$.
$endgroup$
– Chris Culter
Dec 20 '18 at 9:57












$begingroup$
Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 9:59






$begingroup$
Hello @ChrisCulter, yes you are right. I edited my question. Thank you for the revision.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 9:59












2 Answers
2






active

oldest

votes


















2












$begingroup$

You have $m$ and $n$ reversed. The correct expression is:



begin{align}
OC &= OA + mAB \
&=OA+m(OB-OA) \
&=(1-m)OA + mOB\
&=nOA+mOB
end{align}



If you can’t remember where the $m$ and $n$ go in this equation, it’s easy to check the cases $m=0$ and $m=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05



















1












$begingroup$

You're right, except that you have $m$ and $n$ backwards. When $m=0$, we want to get $C=A$, so the correct expression is $C=nA+mB$. This equation expresses $C$ as an affine combination of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You have $m$ and $n$ reversed. The correct expression is:



begin{align}
OC &= OA + mAB \
&=OA+m(OB-OA) \
&=(1-m)OA + mOB\
&=nOA+mOB
end{align}



If you can’t remember where the $m$ and $n$ go in this equation, it’s easy to check the cases $m=0$ and $m=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05
















2












$begingroup$

You have $m$ and $n$ reversed. The correct expression is:



begin{align}
OC &= OA + mAB \
&=OA+m(OB-OA) \
&=(1-m)OA + mOB\
&=nOA+mOB
end{align}



If you can’t remember where the $m$ and $n$ go in this equation, it’s easy to check the cases $m=0$ and $m=1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05














2












2








2





$begingroup$

You have $m$ and $n$ reversed. The correct expression is:



begin{align}
OC &= OA + mAB \
&=OA+m(OB-OA) \
&=(1-m)OA + mOB\
&=nOA+mOB
end{align}



If you can’t remember where the $m$ and $n$ go in this equation, it’s easy to check the cases $m=0$ and $m=1$.






share|cite|improve this answer











$endgroup$



You have $m$ and $n$ reversed. The correct expression is:



begin{align}
OC &= OA + mAB \
&=OA+m(OB-OA) \
&=(1-m)OA + mOB\
&=nOA+mOB
end{align}



If you can’t remember where the $m$ and $n$ go in this equation, it’s easy to check the cases $m=0$ and $m=1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 10:26

























answered Dec 20 '18 at 10:20









bubbabubba

30.8k33188




30.8k33188












  • $begingroup$
    Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05


















  • $begingroup$
    Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05
















$begingroup$
Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 16:05




$begingroup$
Hello @bubba, both your and ChrisCulter answers are excellent, so I voted up them without privileging any.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 16:05











1












$begingroup$

You're right, except that you have $m$ and $n$ backwards. When $m=0$, we want to get $C=A$, so the correct expression is $C=nA+mB$. This equation expresses $C$ as an affine combination of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05
















1












$begingroup$

You're right, except that you have $m$ and $n$ backwards. When $m=0$, we want to get $C=A$, so the correct expression is $C=nA+mB$. This equation expresses $C$ as an affine combination of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05














1












1








1





$begingroup$

You're right, except that you have $m$ and $n$ backwards. When $m=0$, we want to get $C=A$, so the correct expression is $C=nA+mB$. This equation expresses $C$ as an affine combination of $A$ and $B$.






share|cite|improve this answer









$endgroup$



You're right, except that you have $m$ and $n$ backwards. When $m=0$, we want to get $C=A$, so the correct expression is $C=nA+mB$. This equation expresses $C$ as an affine combination of $A$ and $B$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 10:16









Chris CulterChris Culter

21.5k43888




21.5k43888












  • $begingroup$
    Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05


















  • $begingroup$
    Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
    $endgroup$
    – Gennaro Arguzzi
    Dec 20 '18 at 16:05
















$begingroup$
Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 16:05




$begingroup$
Hello @ChrisCulter, both your and bubba answers are excellent, so I voted up them without privileging any.
$endgroup$
– Gennaro Arguzzi
Dec 20 '18 at 16:05


















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