Proving c = acosB + bcosA in Triangle ABC using Vector Method

[Dumbledore211]

Vector Problem please help

Homework Statement

To prove using the vector method for any triangle ABC the statement c=acosB+bcosA holds.


2. Homework Equations [/

A.B=abcosα

The Attempt at a Solution

Frankly, I don't know how to go about it. At first it seemed easy but I just coudn't somehow use the dot product twice to get acosB and bcosA. It would be greatly appreciated if you guy gave me some ideas that would help me this proof..

 

[mfb]

If A and B are vectors, what are cos(A) and cos(B) supposed to be?
a,b,c are the lengths of the sides A, B, C?

That equation looks strange.

 

[Dumbledore211]

No, a,b and c are supposed to be the position vectors of A, B and C. I am first supposed to get the sum of two vectors like for example AB+BC=AC and expand it from here on.

 

[mfb]

Ok, so AB, BC, AC are vectors corresponding to the sides of your triangle.
But what are cos(A) and cos(B)?

 

[Dumbledore211]

Here, acosB= BCcosB and bcosA=ACcosA. acosB and bcosA are also supposed to be two vectors of triangle ABC.

 

[mfb]

I don't think that notation makes any sense.

 

[Dumbledore211]

Okay, Could you give me the starting point of this problem?? CosB is apparently a/c and cosA=b/c.

 

[mfb]

The problem is that I don't even understand your problem.

Can you show the original source? Link, photo, screenshot, whatever?
If that is not possible, please copy the exact problem statement, including all definitions and anything else which belongs to the problem.

CosB is apparently a/c and cosA=b/c.

c is a vector, you cannot divide by vectors.
And even if I replace c by |c|: that would imply c^2=a^2+b^2 for any triangle, which is wrong.

 

[Dumbledore211]

It's actually from my textbook and no descriptive information is given regarding this problem which is why i am at a total loss. The original problem is Prove by vector method in any triangle ABC 1) cosC= a^2+b^2-c^2/2ab
2)c=acosB+bcosA and the only example regarding these problems in the book given is cosA=b^2+c^2-a^2/2bc
BA+AC=BC
or, (BA+AC)(BA+AC)=BC.BC
or, BA^2+AC^2+2BA.AC=BC^2
or, c^2+b^2+2|AC|BA|cos(180-A)
or, c^2+b^2-2bccosA+a^2
or, cosA= b^2+c^2-a^2/2bc
Hope, you wouldn't mind me writing this long post as you yourself told me to present all the definitions and info pertaining to this prob.

 

[mfb]

That post is not long ;).
Apparently A, B, C are the angles in the triangle. Strange, but ok.

cosA=b^2+c^2-a^2/2bc

Just by renaming a,b,c and A,B,C, you get cos(C)= a^2+b^2-c^2/(2ab)

Concerning question 2, did you draw a sketch? Try to mark a*cos(B) on the c-side, beginning at the point where the angle B is. Is the other point on c special in some way?

 

[Dumbledore211]

Sorry but I didn't understand what you meant by your question . small a, b, and c are the position vectors and capital A, B and C are the angles of the triangle. Yes the notations are a bit confusing. Could you just tell me if the dot product will help me in any way in this proof.

 

[mfb]

The dot product can help. But I really suggest to look at the geometric interpretation with a sketch first.

 

[Dumbledore211]

I have sent the sketch in the attachment and see if it's okay.

 

[mfb]

Oh, an additional catch: The equation requires that a,b,c there are the magnitudes of the vectors. With vectors, it is wrong.
$|\vec{c}|=|\vec{a}|\cos(B)+|\vec{b}|\cos(A)$

It is easier to see if you draw a and b such that c=a+b. Just try to modify that equation to get dot products inside.

 

[Dumbledore211]

I think I finally have some idea about solving it. See, if my process is okay or not. I actually have some difficulty using the vector symbols in computers. But, I am first starting off with vectors.

c=a+b
or, BA=BC+CA
or, BA.BA= BC.BA+CA.BA
or, |BA|^2= |BC||BA|cos(180-B) + |CA||BA|cos(180-A)
or, |c|^2= |c||a|cosB +|c||b|cosA
or, |c|= |a|cosB + |b|cosA

 

[mfb]

Without the "180-", but that is good.

 

[Dumbledore211]

Do you think including the "180-" is quite superficial and is it better to exclude it from my solution.

 

[mfb]

I don't see where it should come from, and removing it like you did would switch the sign.

 

[Dumbledore211]

So, thanks for all the help that you have provided in solving this problem and I finally have gained a better understanding that will hopefully help me in solving these problems.