If\( \vec{a}=\vec{b}+\vec{c}\), then is it true that |\(\vec{a}\)|=|\(\vec{b}\)|+|\(\vec{c}\)| ? justify your answer.
In \(△ABC\),let \(\overrightarrow{CB}=\vec{a},\overrightarrow{CA}=\vec{b},\)and \(\overrightarrow{AB}=\vec{c}\)(as shown in the following figure).

Now,by the triangle law of vector addition,we have \(\vec{a}=\vec{b}+\vec{c}\).
It is clearly known that |\(\vec{a}\)|,|\(\vec{b}\)|,and |\(\vec{c}\)|represent the sides of \(△ABC.\)
Also,it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.
∴|\(\vec{a}\)|<|\(\vec{b}\)|+|\(\vec{c}\)|
|Hence,it is not true that |\(\vec{a}\)|=|\(\vec{b}\)|+|\(\vec{c}\)|.
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.
Let \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) be three vectors such that \(\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c}\) and \(\mathbf{a} \times \mathbf{b} \neq 0. Show \;that \;\mathbf{b} = \mathbf{c}\).
The respective values of \( |\vec{a}| \) and} \( |\vec{b}| \), if given \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512 \quad \text{and} \quad |\vec{a}| = 3 |\vec{b}|, \] are: