Proof related to vectors: collinear points, position vectors
$begingroup$
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."
EDIT: mAB+nAB=AB --> m+n=1.
Thank you so much for your time.
vectors education
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
$endgroup$
add a comment |
$begingroup$
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."
EDIT: mAB+nAB=AB --> m+n=1.
Thank you so much for your time.
vectors education
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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."
EDIT: mAB+nAB=AB --> m+n=1.
Thank you so much for your time.
vectors education
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
$endgroup$
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."
EDIT: mAB+nAB=AB --> m+n=1.
Thank you so much for your time.
vectors education
vectors education
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
edited Dec 20 '18 at 9:59
Gennaro Arguzzi
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
asked Dec 20 '18 at 9:50
Gennaro ArguzziGennaro Arguzzi
361314
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $m=1$.
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $B$.
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $m=1$.
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $m=1$.
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $m=1$.
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
$endgroup$
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 $m=1$.
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
edited Dec 20 '18 at 10:26
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
answered Dec 20 '18 at 10:20
bubbabubba
30.8k33188
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $B$.
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $B$.
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $B$.
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
$endgroup$
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 $B$.
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
answered Dec 20 '18 at 10:16
Chris CulterChris Culter
21.5k43888
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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