Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector). 
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b  \end{pmatrix}$ and vector $y$ equal to $\begin{pmatrix} c \\ d \end{pmatrix}$. $||x|| = 7$ would then be, after simplification, $a^2+b^2 = 49$. Similarly, for $||y|| = 11$, after simplification, $c^2+d^2 = 121$.
Then, $||x+y|| = \sqrt{(a+c)^2 + (b+d)^2}$. Expanding gives us $\sqrt{(a^2+b^2) + (c^2+d^2) + 2(ac+bd)} = \sqrt{49+121+2(ac+bd)} = \sqrt{170+2(ac+bd)}$. That is where I was stuck--any hints for the next few steps? 
Part b): $x$ and $y$ are vectors (these are not the same vectors as in part a). If $||x|| = 4, ||y|| = 5, ||x+y|| = 7,$ what is $||2x-3y||$? 
Using the same approach as in part a), where vector $x$ is equal to $\begin{pmatrix} a \\ b  \end{pmatrix}$ and vector $y$ is equal to $\begin{pmatrix} c \\ d \end{pmatrix}$, $a^2+b^2 = 16$ and $c^2+d^2 = 25$. Similarly, for $||x+y||$, after simplification, it equals $41+2(ac+bd) = 49$ -> $ac+bd = 4$. I'm not sure what to do next after this part too. Any hints? 

for the second part it is best to use vectors
$|\vec{x} + \vec{y}|^2 = (\vec{x} + \vec{y})\cdot ( \vec{x} + \vec{y})=
|\vec{x}|^2 + |\vec{y}|^2 +2\vec{x} \cdot \vec{y}$
solve the equation 
$|7|^2 = 
|5|^2 + |4|^2 +2(\vec{x} \cdot \vec{y})$
for $\vec{x} \cdot \vec{y}$
which you can use to evaluate 
$|2\vec{x} -3 \vec{y}|^2 =
4|\vec{x}|^2 + 9|\vec{y}|^2 -6 (\vec{x} \cdot \vec{y})$
