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
2,4441825
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f818798%2ffinding-vectors-in-terms-of-other-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
2,4441825
$endgroup$
$begingroup$
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
2,4441825
$endgroup$
$begingroup$
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
2,4441825
$endgroup$
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
2,4441825
edited Jun 3 '14 at 23:48
answered Jun 3 '14 at 4:27
FabienFabien
2,4441825
answered Jun 3 '14 at 4:27
FabienFabien
2,4441825
answered Jun 3 '14 at 4:27
FabienFabien
2,4441825
2,4441825
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f818798%2ffinding-vectors-in-terms-of-other-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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