If $\vec a = \vec b + \vec c$, then is it true that $|\vec a| = |\vec b| + |\vec c|$? Justify your answer.

We have $\vec a = \vec b + \vec c$ $\therefore \,\,\,|\vec a| = |\vec b + \vec c|$

Squaring, $|\vec a{|^2} = |\vec b + \vec c{|^2}$

$= (\vec b + \vec c) \cdot (\vec b + \vec c) = \vec b \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec b + \vec c \cdot \vec c$

$= |\vec b{|^2} + 2\vec b \cdot \vec c + |\vec c{|^2}$

$= |\vec b{|^2} + |\vec c{|^2} + 2|\vec b||\vec c|\cos \theta$

where $\theta$ is the angle between $\vec b$ and $\vec c$.

When $\theta = {0^\circ }$,

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

$\Rightarrow |\vec a| = |\vec b| + |\vec c|$. When $\theta \ne {0^\circ }$,

then $|\vec a| \ne |\vec b| + |\vec c|$
$\Rightarrow \left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|.$

When $\theta \ne 0^\circ$,

then $\left| {\overrightarrow a } \right| \ne \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|$