# Prove the cosine rule using vectors

Printable View

• Mar 1st 2013, 09:27 PM
iamapineapple
Prove the cosine rule using vectors
Yr 12 Specialist Mathematics:

Triangle ABC where (these are vectors):
AB = a
BC = b
CA = c
such that a + b = -c

Prove the cosine rule, |c|2= |a|2 + |b|2 -2 |a|.|b| cosB using vectors

So far, I've been able to derive |c|2= |a|2 + |b|2 +2 |a|.|b| cosB, with a positive not a negative. I used dot product rules where c.c = |(-a-b)2|cosB. I'm a bit lost, and could really use some help on how to get the answer! :D

And this is my first time on these forums XD Could someone please tell me if there are symbols we can use? For pi, square root, things like that?

Much appreciation
-iamapineapple
• Mar 1st 2013, 11:56 PM
Prove It
Re: Prove the cosine rule using vectors
• Mar 1st 2013, 11:59 PM
iamapineapple
Re: Prove the cosine rule using vectors
Thanks and all, but that's the question I have to answer. "Prove it using vectors"......
• Mar 2nd 2013, 03:55 AM
Plato
Re: Prove the cosine rule using vectors
Quote:

Originally Posted by iamapineapple
Yr 12 Specialist Mathematics:
Triangle ABC where (these are vectors):
AB = a
BC = b
CA = c
such that a + b = -c
Prove the cosine rule, |c|2= |a|2 + |b|2 -2 |a|.|b| cosB using vectors

You have several errors there.

From the given, $\cos(B)=\frac{-\vec{a}\cdot\vec{b}}{\|-\vec{a}\|\|\vec{b}\|}$.

$\|\vec{c}\|^2=(-\vec{a}-\vec{b})\cdot(-\vec{a}-\vec{b})=\|(\vec{a}\|^2+\|(\vec{b}\|^2+2\vec{a} \cdot \vec{b}$