Show that given a point P belongs to a triangel if and only if..

[Randyyy]

Show that a point P is on a triangel ABC if and only if [MATH]\vec{AP}=t\vec{AB}+s\vec{AC} [/MATH] where [MATH] s,t \geq 0, s+t \leq 1[/MATH]
Hey, I have no idea how to prove this or even where to begin. I tried drawing a triangle with the labels ABC and drew out a random point P on the triangle but I don´t see how I would satisfy the proof. My picture is a bit boring since it doesn´t really show anything special but I guess it´s better to include it than to not include it!

Thanks in advance.

 

[lev888]

Show that a point P is on a triangel ABC if and only if [MATH]\vec{AP}=t\vec{AB}+s\vec{AC} [/MATH] where [MATH] s,t \geq 0, s+t \leq 1[/MATH]
Hey, I have no idea how to prove this or even where to begin. I tried drawing a triangle with the labels ABC and drew out a random point P on the triangle but I don´t see how I would satisfy the proof. My picture is a bit boring since it doesn´t really show anything special but I guess it´s better to include it than to not include it!

Thanks in advance.


View attachment 25011

I would draw geometric representation of the vector sum.
You need 2 proofs here (that's what "if and only if" means). Can you state what needs to be proven in each one?

 

[Randyyy]

Yes, I need to prove it both ways. I have to show that [MATH]\vec{AP} \implies t\vec{AB}+s\vec{AC}[/MATH] and that [MATH]t\vec{AB}+s\vec{AC}\implies \vec{AP}[/MATH]. I was thinking of maybe a vector sum but I got stuck on how to treat the scalars in front of my two vectors. I´m guessing that [MATH]\vec{AB}[/MATH] would be the hypothenuse. So in essence that [MATH]t\vec{AB}+s\vec{AC} [/MATH] with the hypothenuse [MATH]t\vec{AB}+s\vec{AC}[/MATH] but I don´t see how I would show that it somehow implies [MATH]\vec{AB}[/MATH]. I guess s and t are scalars and they only determine the length so I could just draw arbitrary vectors.

 

[Randyyy]

Yes, I need to prove it both ways. I have to show that [MATH]\vec{AP} \implies t\vec{AB}+s\vec{AC}[/MATH] and that [MATH]t\vec{AB}+s\vec{AC}\implies \vec{AP}[/MATH]. I was thinking of maybe a vector sum but I got stuck on how to treat the scalars in front of my two vectors. I´m guessing that [MATH]\vec{AB}[/MATH] would be the hypothenuse. So in essence that [MATH]t\vec{AB}+s\vec{AC} [/MATH] with the hypothenuse [MATH]t\vec{AB}+s\vec{AC}[/MATH] but I don´t see how I would show that it somehow implies [MATH]\vec{AB}[/MATH]. I guess s and t are scalars and they only determine the length so I could just draw arbitrary vectors.

View attachment 25016

Perhaps like so? I put A as my origin and put a random point inside the plane and It clearly shows (if what I have written is correct) that AP=sAB+tAC.

 

[lev888]

Yes, I need to prove it both ways. I have to show that →AP⟹t→AB+s→AC\displaystyle \vec{AP} \implies t\vec{AB}+s\vec{AC} and that t→AB+s→AC⟹→AP\displaystyle t\vec{AB}+s\vec{AC}\implies \vec{AP}.

I have no idea what you mean by the above. What you need to prove is:
1. If point P lies within the triangle ABC, then AP=.....
2. If AP=....., then point P lies within the triangle ABC

 

[Randyyy]

I have no idea what you mean by the above. What you need to prove is:
1. If point P lies within the triangle ABC, then AP=.....
2. If AP=....., then point P lies within the triangle ABC

Yeah I realized myself that my pictures makes no sense at all.

1. If point P lies within the triangle ABC, then AP=[MATH]t\vec{AB}+s\vec{AC}[/MATH]2. If AP=[MATH]t\vec{AB}+s\vec{AC}[/MATH] then point P lies within the triangle ABC.

 

[lev888]

Yeah I realized myself that my pictures makes no sense at all.

1. If point P lies within the triangle ABC, then AP=[MATH]t\vec{AB}+s\vec{AC}[/MATH]2. If AP=[MATH]t\vec{AB}+s\vec{AC}[/MATH] then point P lies within the triangle ABC.

I was commenting on your interpretation of the 2 proofs, not the diagram. And yes, the diagram is lacking. First, I think it should include the whole triangle. Second, if you are using the "Parallelogram Method" for vector addition, then we should see a parallelogram. Third, are those coordinates of point P in parentheses? In any case, it does look like if P is within ABC, then AP can be expressed as a sum of those vectors. But we also need to show that the restrictions on s and t are satisfied.
And then you need to prove the 2nd statement.