If a+b+c=0 and the magnitudes |a| , |b| , |c| are known…

If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and the magnitudes $|\vec{a}|$, $|\vec{b}|$, $|\vec{c}|$ are known, find $\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}$.

Step 1: Square the condition

\[ (\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = \vec 0\cdot\vec 0 = 0 \]

Expanding (nine terms) and using commutativity and $\vec u\cdot\vec u=|\vec u|^2$: \[|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}) = 0\]

\[\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c} = -\frac{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}{2}\]

Step 2: The individual products

From $\vec c=-(\vec a+\vec b)$ we get

\[ |\vec c|^2 = |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2, \] and cyclically: \[\vec{a}\cdot\vec{b} = \frac{|\vec{c}|^2-|\vec{a}|^2-|\vec{b}|^2}{2}, \quad \vec{a}\cdot\vec{c} = \frac{|\vec{b}|^2-|\vec{a}|^2-|\vec{c}|^2}{2}, \quad \vec{b}\cdot\vec{c} = \frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2}.\]

Adding them gives the result again $\checkmark$.

Step 3: Geometric meaning

Placed head-to-tail, the three vectors form a closed triangle with sides $|\vec a|,|\vec b|,|\vec c|$. The formulas above are the law of cosines. Note that the angle between the vectors $\vec a$ and $\vec b$ (tails together) is the exterior angle $180^\circ-\varphi$, where $\varphi$ is the interior angle of the triangle at the common vertex; that is why the signs look reversed compared with the usual form $c^2=a^2+b^2-2ab\cos\varphi$.

Step 4: Example

For $|\vec a|=3$, $|\vec b|=4$, $|\vec c|=5$: $\vec a\cdot\vec b+\vec a\cdot\vec c+\vec b\cdot\vec c = -\frac{9+16+25}{2} = -25$.

\[ \vec{a}\cdot\vec{b}=\frac{25-9-16}{2}=0\ (\vec a\perp\vec b), \qquad \vec{a}\cdot\vec{c}=-9, \qquad \vec{b}\cdot\vec{c}=-16, \qquad 0-9-16=-25 \;\checkmark \]