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 \(\displaystyle \lambda + \mu + \nu = 1\).

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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 \(\displaystyle \lambda + \mu + \nu = 1\).

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Please show us what you have tried and

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 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.Please show us what you have tried andexactlywhere you are stuck.

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

Please share your work/thoughts about this problem

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Please share your efforts.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.

I want to prove:

If A, B, C are the vertices of a triangle and X is a point

Yes

I want to prove:

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

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Please share your efforts.

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.

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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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.

Are these the same question?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..?

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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I got from that method ..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.

Great. Well done.I got from that method ..

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

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..

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!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!)

Have you got it done then? Post your work if there is a problem.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!