Finding vectors in terms of other vectors
$begingroup$
Suppose that O, A and B are three non-collinear points in a plane.
Let $vec {OC}:=vec {OB}-vec{2OA}$, and $vec {OE} :=vec {-OA}$
Express $vec {OM}$ in terms of the vectors $vec {OA}$ and $vec {OB}$ where M is the point of intersection of the line through O and C and the line through B and E.
I dont know how to come up with 2 equations for OM to solve. I would need a step by step answer if that is possible, any takers?
vectors
asked Jun 3 '14 at 3:55
user148615user148615
116
$endgroup$
add a comment |
$begingroup$
Suppose that O, A and B are three non-collinear points in a plane.
Let $vec {OC}:=vec {OB}-vec{2OA}$, and $vec {OE} :=vec {-OA}$
Express $vec {OM}$ in terms of the vectors $vec {OA}$ and $vec {OB}$ where M is the point of intersection of the line through O and C and the line through B and E.
I dont know how to come up with 2 equations for OM to solve. I would need a step by step answer if that is possible, any takers?
vectors
asked Jun 3 '14 at 3:55
user148615user148615
116
$endgroup$
1
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04
add a comment |
$begingroup$
Suppose that O, A and B are three non-collinear points in a plane.
Let $vec {OC}:=vec {OB}-vec{2OA}$, and $vec {OE} :=vec {-OA}$
Express $vec {OM}$ in terms of the vectors $vec {OA}$ and $vec {OB}$ where M is the point of intersection of the line through O and C and the line through B and E.
I dont know how to come up with 2 equations for OM to solve. I would need a step by step answer if that is possible, any takers?
vectors
asked Jun 3 '14 at 3:55
user148615user148615
116
$endgroup$
Suppose that O, A and B are three non-collinear points in a plane.
Let $vec {OC}:=vec {OB}-vec{2OA}$, and $vec {OE} :=vec {-OA}$
Express $vec {OM}$ in terms of the vectors $vec {OA}$ and $vec {OB}$ where M is the point of intersection of the line through O and C and the line through B and E.
I dont know how to come up with 2 equations for OM to solve. I would need a step by step answer if that is possible, any takers?
vectors
vectors
asked Jun 3 '14 at 3:55
user148615user148615
116
asked Jun 3 '14 at 3:55
user148615user148615
116
edited Jun 3 '14 at 6:57
user148615
asked Jun 3 '14 at 3:55
user148615user148615
116
asked Jun 3 '14 at 3:55
user148615user148615
116
asked Jun 3 '14 at 3:55
user148615user148615
116
116
1
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04
add a comment |
1
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04
1
1
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $vec{OM}$:
$$vec{OM}=alphavec{OC}=alpha(vec{OB}-2vec{OA})$$
$$vec{OM}=vec{OE}+vec{EM}=vec{OE}+betavec{EB}=vec{OE}+beta(vec{EO}+vec{OB})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$alpha(vec{OB}-2vec{OA})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$(alpha-beta)vec{OB} + (-2alpha+1-beta)vec{OA}=vec{0}$$
so
$$begin{cases} alpha-beta=0 \ -2alpha+1-beta=0 end{cases}$$
so
$$begin{cases} alpha=beta \ alpha=frac13 end{cases}$$
so $$vec{OM}=frac13(vec{OB}-2vec{OA})$$
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
$endgroup$
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
add a comment |
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$begingroup$
Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $vec{OM}$:
$$vec{OM}=alphavec{OC}=alpha(vec{OB}-2vec{OA})$$
$$vec{OM}=vec{OE}+vec{EM}=vec{OE}+betavec{EB}=vec{OE}+beta(vec{EO}+vec{OB})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$alpha(vec{OB}-2vec{OA})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$(alpha-beta)vec{OB} + (-2alpha+1-beta)vec{OA}=vec{0}$$
so
$$begin{cases} alpha-beta=0 \ -2alpha+1-beta=0 end{cases}$$
so
$$begin{cases} alpha=beta \ alpha=frac13 end{cases}$$
so $$vec{OM}=frac13(vec{OB}-2vec{OA})$$
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
$endgroup$
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
add a comment |
$begingroup$
Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $vec{OM}$:
$$vec{OM}=alphavec{OC}=alpha(vec{OB}-2vec{OA})$$
$$vec{OM}=vec{OE}+vec{EM}=vec{OE}+betavec{EB}=vec{OE}+beta(vec{EO}+vec{OB})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$alpha(vec{OB}-2vec{OA})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$(alpha-beta)vec{OB} + (-2alpha+1-beta)vec{OA}=vec{0}$$
so
$$begin{cases} alpha-beta=0 \ -2alpha+1-beta=0 end{cases}$$
so
$$begin{cases} alpha=beta \ alpha=frac13 end{cases}$$
so $$vec{OM}=frac13(vec{OB}-2vec{OA})$$
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
$endgroup$
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
add a comment |
$begingroup$
Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $vec{OM}$:
$$vec{OM}=alphavec{OC}=alpha(vec{OB}-2vec{OA})$$
$$vec{OM}=vec{OE}+vec{EM}=vec{OE}+betavec{EB}=vec{OE}+beta(vec{EO}+vec{OB})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$alpha(vec{OB}-2vec{OA})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$(alpha-beta)vec{OB} + (-2alpha+1-beta)vec{OA}=vec{0}$$
so
$$begin{cases} alpha-beta=0 \ -2alpha+1-beta=0 end{cases}$$
so
$$begin{cases} alpha=beta \ alpha=frac13 end{cases}$$
so $$vec{OM}=frac13(vec{OB}-2vec{OA})$$
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
$endgroup$
Since M belongs to $(OC)$ and $(BE)$, then you have two ways to consider $vec{OM}$:
$$vec{OM}=alphavec{OC}=alpha(vec{OB}-2vec{OA})$$
$$vec{OM}=vec{OE}+vec{EM}=vec{OE}+betavec{EB}=vec{OE}+beta(vec{EO}+vec{OB})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$alpha(vec{OB}-2vec{OA})=-vec{OA}+beta(vec{OA}+vec{OB})$$
so
$$(alpha-beta)vec{OB} + (-2alpha+1-beta)vec{OA}=vec{0}$$
so
$$begin{cases} alpha-beta=0 \ -2alpha+1-beta=0 end{cases}$$
so
$$begin{cases} alpha=beta \ alpha=frac13 end{cases}$$
so $$vec{OM}=frac13(vec{OB}-2vec{OA})$$
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
edited Jun 3 '14 at 23:48
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
answered Jun 3 '14 at 4:27
FabienFabien
2,4741825
2,4741825
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
add a comment |
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
Hi Fabien, Thank you for answering, just wondering how you managed to deduce the first 2 equations? Also I am unsure what is meant by the first and second projection and also why is it required and how you came about finding the alpha values?
$endgroup$
– user148615
Jun 3 '14 at 10:04
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
@user148615, I've edited. Feel free to post another comment if it's not clear enough ;)
$endgroup$
– Fabien
Jun 3 '14 at 12:05
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
Hi again Fabien, thank you for the reply, just wondering how you managed to conclude that, on the second line of equations, the vector EB is equal to OA+OB? Is it because the two vectors are parallel?
$endgroup$
– user148615
Jun 3 '14 at 23:45
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
@user148615, consider $vec{EB}=vec{EO}+vec{OB}$
$endgroup$
– Fabien
Jun 3 '14 at 23:48
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
$begingroup$
Ah sorry! I have always learnt vectors to be in the form EB = OB - OE but I realised that E is actually in front of the O which makes EO=OA :) Thankyou Fabien!
$endgroup$
– user148615
Jun 3 '14 at 23:56
add a comment |
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1
$begingroup$
Could you post the diagram you drew?
$endgroup$
– Carser
Jun 3 '14 at 4:01
$begingroup$
Assigning O as origin might simplify the problem
$endgroup$
– GTX OC
Jun 3 '14 at 4:01
$begingroup$
math.stackexchange.com/questions/818899/…
$endgroup$
– Gerry Myerson
Jun 3 '14 at 7:23
$begingroup$
Hi sorry, I tried to add a comment but could not, then I changed from google chrome to internet explorer and it worked, very sorry.
$endgroup$
– user148615
Jun 3 '14 at 10:04