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I if want to prove the cosine rule with the vector summation (such as I have mentioned in the picture) I start with the $\vec{a}+\vec{b}=\vec{c}$

$\vec{a}=\vec{c}-\vec{b}$ then $a^2=\vec{a}.\vec{a}=(\vec{c}-\vec{b}).(\vec{c}-\vec{b})=b^2+c^2-2bcCos(A)$ which is completely correct.

enter image description here

If I change the vector such as this picture

enter image description here

I tried to write $\vec{a}+\vec{b}=-\vec{c}$
$\vec{a}=-\vec{c}-\vec{b}$ $a^2=\vec{a}.\vec{a}=(\vec{c}+\vec{b}).(\vec{c}+\vec{b})=b^2+c^2+2bcCos(A)$

The positive sign before cosine is my problem.

tinlyx
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John Jin
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1 Answers1

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The angle $\theta$ you meant from the identity $\vec u\cdot \vec v =uv\cos (\theta)$ is to be measured between the two vectors when the tail is at the same point. In the figure you have shown the angle $A$ is not what you use. You need $\pi -A$. From this, you reduct to angle and would recover the missing minus sign.

Korn Kruaykitanon
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