# Proving a good result in vectors in 3D geometry

#### Rover

##### New member
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

##### Super Moderator
Staff member
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:

#### Rover

##### New member
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:

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

##### Super Moderator
Staff member
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.

#### lex

##### Full Member
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

##### New member
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

Staff member

#### lex

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

#### Rover

##### New member
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.

Last edited by a moderator:

#### lex

##### Full Member
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.

Last edited:

#### Rover

##### New member
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

##### New member
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

##### New member
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

##### Full Member
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

##### New member
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

##### Full Member
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.