Proving a good result in vectors in 3D geometry

Rover

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I want to Prove:
If A,B,C are the vertices of a triangle and X is a point in the plane, you can write X=\(\displaystyle \lambda A + \mu B + \nu \) for unique scalars \(\displaystyle \lambda,\mu,\nu \) with \(\displaystyle \lambda + \mu + \nu = 1\).
 

Subhotosh Khan

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I want to Prove:
If A,B,C are the vertices of a triangle and X is a point in the plane, you can write X=\(\displaystyle \lambda A + \mu B + \nu \) for unique scalars \(\displaystyle \lambda,\mu,\nu \) with \(\displaystyle \lambda + \mu + \nu = 1\).
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 

Rover

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Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
I started by taking a any X with \vector{r} and tried to relate by using relates knowledge, but I am unable to proceed from long time, so I want some help.
 

Subhotosh Khan

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I started by taking a any X with \vector{r} and tried to relate by using relates knowledge, but I am unable to proceed from long time, so I want some help.
Please share your efforts.
 

lex

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I assume you mean -

I want to prove:
If A, B, C are the vertices of a triangle and X is a point in the plane of the triangle, then you can write: \(\displaystyle \boldsymbol{x}=λ\boldsymbol{a}+μ\boldsymbol{b}+\nu \boldsymbol{c},\hspace1ex \text{ where }\lambda,\; \mu,\; \nu \; \text{ are scalars: } \lambda + \mu+\nu=1 \text{ and }\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{x} \text{ are the position vectors of the points A, B, C, X respectively}.\)
 

Rover

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I assume you mean -

I want to prove:
If A, B, C are the vertices of a triangle and X is a point in the plane of the triangle, then you can write: \(\displaystyle \boldsymbol{x}=λ\boldsymbol{a}+μ\boldsymbol{b}+\nu \boldsymbol{c},\hspace1ex \text{ where }\lambda,\; \mu,\; \nu \; \text{ are scalars: } \lambda + \mu+\nu=1 \text{ and }\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{x} \text{ are the position vectors of the points A, B, C, X respectively}.\)
Yes
 

lex

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Can you give an indication of what you know, what course you are following or what textbook you are using?
 

Rover

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Okay, so I am doing a question from 3D geometry and the question states as follows:A point O is a centre of circle circumscribed about a triangle ABC.Then, \(\displaystyle \vec{OA}sin2A+\vec{OB}sin2B+\vec{OB}sin2C \) is equal to..?
And for this question I tried a lot , first I started by computing the areas of OAB,OBC,OCA in terms of sin2A,sin2B,sin2C , then I tried to relate vector OA, OB,OC to it by cross product but it didn't worked ...
And if I just solve the question by taking example like if it's a equilateral triangle or so then I get answer 0, but I want to prove it.
 
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lex

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I want to Prove:
If A,B,C are the vertices of a triangle and X is a point in the plane, you can write X=λA+μB+νλA+μB+ν\displaystyle \lambda A + \mu B + \nu for unique scalars λ,μ,νλ,μ,ν\displaystyle \lambda,\mu,\nu with λ+μ+ν=1λ+μ+ν=1\displaystyle \lambda + \mu + \nu = 1.
Okay, so I am doing a question from 3D geometry and the question states as follows:A point O is a centre of circle circumscribed about a triangle ABC.Then, \vec{OA}sin2A+\vec{OB}sin2B+\vec{OB}sin2C is equal to..?
Are these the same question?

Concerning the second question. Are you sure this is what it says? Please check it.

Is it not:\(\displaystyle \hspace2ex \vec{OA}\sin2A+\vec{OB}\sin2B+\vec{OC}\sin2C\) is equal to..?

In which case try getting the scalar product of the expression with \(\displaystyle \vec{OA}\)
and the scalar product of the expression with \(\displaystyle \vec{OB}\)
and see what you can conclude from that.
 
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Rover

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Are these the same question?

Concerning the second question. Are you sure this is what it says? Please check it.

Is it not:\(\displaystyle \hspace2ex \vec{OA}\sin2A+\vec{OB}\sin2B+\vec{OC}\sin2C\) is equal to..?

In which case try getting the scalar product of the expression with \(\displaystyle \vec{OA}\)
and the scalar product of the expression with \(\displaystyle \vec{OB}\)
and see what you can conclude from that.
I got from that method ..
 

Rover

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But, first I want to prove the first question and with that considering as general result, I think have other method in mind, so I want help in proving my first question.
 

Rover

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Great. Well done.
No, I got already before putting on this site..

But, first I want to prove the first question and with that considering as general result, I think have other method in mind, so I want help in proving my first question.(I wrote this 3 hrs ago, but it shows,"This message is awaiting moderator approval, and is invisible to normal visitors.")
I want help for my question I posted on site, it's main for me..
 

lex

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If A, B, C are the vertices of a triangle and X is a point in the plane of the triangle, then you can write: \(\displaystyle \boldsymbol{x}=λ\boldsymbol{a}+μ\boldsymbol{b}+\nu \boldsymbol{c},\hspace1ex \text{ where }\lambda,\; \mu,\; \nu \; \text{ are scalars: } \lambda + \mu+\nu=1 \text{ and }\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{x} \text{ are the position vectors of the points A, B, C, X respectively}.\)

Do you know that any vector in the plane can be written uniquely as a linear combination of 2 non-parallel vectors in the plane?
If so, assuming A, B, C form a (non-degenerate) triangle, then \(\displaystyle \vec{CA}=\boldsymbol{a}-\boldsymbol{c}\), \(\displaystyle \vec{CB}=\boldsymbol{b}-\boldsymbol{c}\) are two non-parallel vectors in the plane.
\(\displaystyle \vec{CX}=\boldsymbol{x}-\boldsymbol{c}\), being a vector in the plane, can be written uniquely as \(\displaystyle \hspace2ex λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})\)
i.e. \(\displaystyle \hspace1ex \exists! \hspace2ex \lambda \text{, } \mu: \hspace2ex \boldsymbol{x}-\boldsymbol{c}= λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})\)
Now re-arrange that to get the form you are seeking and you should be ok.
(I hope I'm not begging the question!)
 

Rover

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If A, B, C are the vertices of a triangle and X is a point in the plane of the triangle, then you can write: \(\displaystyle \boldsymbol{x}=λ\boldsymbol{a}+μ\boldsymbol{b}+\nu \boldsymbol{c},\hspace1ex \text{ where }\lambda,\; \mu,\; \nu \; \text{ are scalars: } \lambda + \mu+\nu=1 \text{ and }\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}, \boldsymbol{x} \text{ are the position vectors of the points A, B, C, X respectively}.\)

Do you know that any vector in the plane can be written uniquely as a linear combination of 2 non-parallel vectors in the plane?
If so, assuming A, B, C form a (non-degenerate) triangle, then \(\displaystyle \vec{CA}=\boldsymbol{a}-\boldsymbol{c}\), \(\displaystyle \vec{CB}=\boldsymbol{b}-\boldsymbol{c}\) are two non-parallel vectors in the plane.
\(\displaystyle \vec{CX}=\boldsymbol{x}-\boldsymbol{c}\), being a vector in the plane, can be written uniquely as \(\displaystyle \hspace2ex λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})\)
i.e. \(\displaystyle \hspace1ex \exists! \hspace2ex \lambda \text{, } \mu: \hspace2ex \boldsymbol{x}-\boldsymbol{c}= λ(\boldsymbol{a}-\boldsymbol{c})+μ(\boldsymbol{b}-\boldsymbol{c})\)
Now re-arrange that to get the form you are seeking and you should be ok.
(I hope I'm not begging the question!)
Yes I know, "any vector in the plane can be written uniquely as a linear combination of 2 non-parallel vectors in the plane". thanks!
 

lex

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Yes I know, "any vector in the plane can be written uniquely as a linear combination of 2 non-parallel vectors in the plane". thanks!
Have you got it done then? Post your work if there is a problem.
 
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