If ${a}$ and ${b}$ are vectors such that $\|a\| = 4$, $\|b\| = 5$, and $\|a + b\| = 7$, then find $\|2a - 3b\|$. If ${a}$ and ${b}$ are vectors such that $\|a\| = 4$, $\|b\| = 5$, and $\|a + b\| = 7$, then find $\|2a - 3b\|$.    
sidenote I'm wondering if $\|a - b\|^2$ is a thing because I know $\| a + b \|^2$ is a thing.
I'm a little confused on how to start. Like, am I suppose magically make $\|a\|$ into $2a$?? Or try to start a path with the number 4, 5, and 7??
Help with be greatly appreciated!!!

For this, $||x||^2$ is equivalent to $(x \cdot x)$.  
$$\begin{align}
(7^2) & = ||a + b||^2 \\
 & = (a + b)\cdot(a+b) \\
 & = a \cdot a + b \cdot a + a \cdot b + b \cdot b\\
 & = a \cdot a + 2 (a \cdot b) + b \cdot b \\
 & = ||a||^2 + 2 (a \cdot b) + ||b||^2\\
(7^2) & = (4^2) + 2 (a \cdot b) + (5^2)\\
49 & = 16 + 2 (a \cdot b) + 25
\end{align}$$
So $a \cdot b = 4$
$$\begin{align}
||2a - 3b||^2 & = (2a - 3b) \cdot (2a - 3b)\\
& = (2a \cdot 2a) + (-3b \cdot 2a) + (2a \cdot (-3b)) + ((-3b) \cdot (-3b))\\
& = 4(a \cdot a) - 12 (a \cdot b) + 9 (b \cdot b)\\
& = 4 ||a||^2 - 12 (a \cdot b) + 9 ||b||^2\\
& = 4 (4^2) - 12 (4) + 9(5^2)\\
& = 64 - 48 + 225\\
& = 241
\end{align}$$
So $||2a - 3b|| = \sqrt{241}$
