Let $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ be three vectors such that $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$.
If $\left|\overrightarrow{u}\right| = 3$, $\left|\overrightarrow{v}\right| = 4$ and $\left|\overrightarrow{w}\right| = 5$, then find $\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}$
Let
$\alpha$ be the angle between $\overrightarrow{u}$ and $\overrightarrow{v}$
$\beta$ be the angle between $\overrightarrow{v}$ and $\overrightarrow{w}$
$\gamma$ be the angle between $\overrightarrow{w}$ and $\overrightarrow{u}$
Given: $\left|\overrightarrow{u}\right| = 3$, $\left|\overrightarrow{v}\right| = 4$ and $\left|\overrightarrow{w}\right| = 5$
Given: $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$
$\implies$ $\overrightarrow{u} + \overrightarrow{v} = - \overrightarrow{w}$
Then, $\;$ $\left(\overrightarrow{u} + \overrightarrow{v}\right)^2 = \left(- \overrightarrow{w}\right)^2$
i.e. $\;$ $\left|\overrightarrow{u}\right|^2 + \left|\overrightarrow{v}\right|^2 + 2 \left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right| \cos \alpha = \left|w\right|^2$
i.e. $\;$ $\left(3\right)^2 + \left(4\right)^2 + 2 \times 3 \times 4 \times \cos \alpha = \left(5\right)^2$
i.e. $\;$ $\cos \alpha = \dfrac{25 - 9 - 16}{24} = 0$
$\because$ $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$
$\implies$ $\overrightarrow{v} + \overrightarrow{w} = - \overrightarrow{u}$
Then, $\;$ $\left(\overrightarrow{v} + \overrightarrow{w}\right)^2 = \left(- \overrightarrow{u}\right)^2$
i.e. $\;$ $\left|\overrightarrow{v}\right|^2 + \left|\overrightarrow{w}\right|^2 + 2 \left|\overrightarrow{v}\right| \left|\overrightarrow{w}\right| \cos \beta = \left|u\right|^2$
i.e. $\;$ $\left(4\right)^2 + \left(5\right)^2 + 2 \times 4 \times 5 \times \cos \beta = \left(3\right)^2$
i.e. $\;$ $\cos \beta = \dfrac{9 - 16 - 25}{40} = \dfrac{-32}{40} = \dfrac{-4}{5}$
$\because$ $\;$ $\overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} = \overrightarrow{0}$
$\implies$ $\overrightarrow{u} + \overrightarrow{w} = - \overrightarrow{v}$
Then, $\;$ $\left(\overrightarrow{u} + \overrightarrow{w}\right)^2 = \left(- \overrightarrow{v}\right)^2$
i.e. $\;$ $\left|\overrightarrow{u}\right|^2 + \left|\overrightarrow{w}\right|^2 + 2 \left|\overrightarrow{u}\right| \left|\overrightarrow{w}\right| \cos \gamma = \left|v\right|^2$
i.e. $\;$ $\left(3\right)^2 + \left(5\right)^2 + 2 \times 3 \times 5 \times \cos \gamma = \left(4\right)^2$
i.e. $\;$ $\cos \gamma = \dfrac{16 - 9 - 25}{30} = \dfrac{-18}{30} = \dfrac{-3}{5}$
Now, $\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{w} \cdot \overrightarrow{u}$
$= \left|\overrightarrow{u}\right| \left|\overrightarrow{v}\right| \cos \alpha + \left|\overrightarrow{v}\right| \left|\overrightarrow{w}\right| \cos \beta + \left|\overrightarrow{w}\right| \left|\overrightarrow{u}\right| \cos \gamma$
$= 3 \times 4 \times
0 + 4 \times 5 \times \left(\dfrac{-4}{5}\right) + 5 \times 3 \times \left(\dfrac{-3}{5}\right) = - 25$