Answer
$|\vec{u}|=\sqrt{13}$
$|\vec{v}|=\sqrt{5}$
$|2\vec{u}|=2\sqrt{13}$
$|\frac{1}{2}\vec{v}|=\dfrac{\sqrt{5}}{2}$
$|\vec{u}+\vec{v}|=\sqrt{2}$
$|\vec{u}-\vec{v}|=\sqrt{34}$
$|\vec{u}|-|\vec{v}|=\sqrt{13}-\sqrt{5}$
Work Step by Step
$\vec{u}=-2\vec{i}+3\vec{j}$ $;$ $\vec{v}=\vec{i}-2\vec{j}$
To obtain the magnitude of a vector $\vec{v}=\langle a_{1},a_{2}\rangle$, apply the formula $|\vec{v}|=\sqrt{a_{1}^{2}+a_{2}^{2}}$
$|\vec{u}|$
For the vector $\vec{u}$, $a_{1}=-2$ and $a_{2}=3$
$|\vec{u}|=\sqrt{(-2)^{2}+3^{2}}=\sqrt{4+9}=\sqrt{13}$
$|\vec{v}|$
For the vector $\vec{v}$, $a_{1}=1$ and $a_{2}=-2$
$|\vec{v}|=\sqrt{1^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5}$
$|2\vec{u}|$
Multiply both components of $\vec{u}$ by $2$:
$2\vec{u}=2(-2\vec{i}+3\vec{j})=-4\vec{i}+6\vec{j}$
For the vector $2\vec{u}$, $a_{1}=-4$ and $a_{2}=6$:
$|2\vec{u}|=\sqrt{(-4)^{2}+6^{2}}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$
$|\frac{1}{2}\vec{v}|$
Multiply both components of $\vec{v}$ by $\dfrac{1}{2}$:
$\dfrac{1}{2}\vec{v}=\dfrac{1}{2}(\vec{i}-2\vec{j})=\dfrac{1}{2}\vec{i}-\vec{j}$
For the vector $|\frac{1}{2}\vec{v}|$, $a_{1}=\dfrac{1}{2}$ and $a_{2}=-1$:
$|\frac{1}{2}\vec{v}|=\sqrt{\Big(\dfrac{1}{2}\Big)^{2}+(-1)^{2}}=\sqrt{\dfrac{1}{4}+1}=\sqrt{\dfrac{5}{4}}=\dfrac{\sqrt{5}}{2}$
$|\vec{u}+\vec{v}|$
Add the vectors $\vec{u}$ and $\vec{v}$ together and simplify:
$\vec{u}+\vec{v}=-2\vec{i}+3\vec{j}+\vec{i}-2\vec{j}=(-2+1)\vec{i}+(3-2)\vec{j}=-\vec{i}+\vec{j}$
For the vector $\vec{u}+\vec{j}$, $a_{1}=-1$ and $a_{2}=1$:
$|\vec{u}+\vec{v}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}$
$|\vec{u}-\vec{v}|$
Subtract vectors $\vec{u}$ and $\vec{v}$ and simplify:
$\vec{u}-\vec{v}=(-2\vec{i}+3\vec{j})-(\vec{i}-2\vec{j})=-2\vec{i}+3\vec{j}-\vec{i}+2\vec{j}=...$
$...=(-2-1)\vec{i}+(3+2)\vec{j}=-3\vec{i}+5\vec{j}$
For the vector $\vec{u}-\vec{v}$, $a_{1}=-3$ and $a_{2}=5$:
$|\vec{u}-\vec{v}|=\sqrt{(-3)^{2}+5^{2}}=\sqrt{9+25}=\sqrt{34}$
$|\vec{u}|-|\vec{v}|$
Subtract the magnitude of $\vec{v}$ from the magnitude of $\vec{u}$. These two magnitudes where found in the first two parts of this problem. $|\vec{u}|=\sqrt{13}$ and $|\vec{v}|=\sqrt{5}$
$|\vec{u}|-|\vec{v}|=\sqrt{13}-\sqrt{5}$