Answer
$|\vec{u}|=\sqrt{101}$
$|\vec{v}|=2\sqrt{2}$
$|2\vec{u}|=2\sqrt{101}$
$|\frac{1}{2}\vec{v}|=\sqrt{2}$
$|\vec{u}+\vec{v}|=\sqrt{73}$
$|\vec{u}-\vec{v}|=\sqrt{145}$
$|\vec{u}|-|\vec{v}|=\sqrt{101}-2\sqrt{2}$
Work Step by Step
$\vec{u}=\langle10,-1\rangle$ $,$ $\vec{v}=\langle-2,-2\rangle$
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 $\vec{u}$, $a_{1}=10$ and $a_{2}=-1$
$|\vec{u}|=\sqrt{10^{2}+(-1)^{2}}=\sqrt{100+1}=\sqrt{101}$
$|\vec{v}|$
For the vector $\vec{v}$, $a_{1}=-2$ and $a_{2}=-2$
$|\vec{v}|=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
$|2\vec{u}|$
Multiply both components of $\vec{u}$ by $2$:
$2\vec{u}=2\langle10,-1\rangle=\langle20,-2\rangle$
For the vector $2\vec{u}$, $a_{1}=20$ and $a_{2}=-2$
$|2\vec{u}|=\sqrt{20^{2}+(-2)^{2}}=\sqrt{400+4}=\sqrt{404}=2\sqrt{101}$
$|\frac{1}{2}\vec{v}|$
Multiply both components of $\vec{v}$ by $\dfrac{1}{2}$:
$\dfrac{1}{2}\vec{v}=\dfrac{1}{2}\langle-2,-2\rangle=\langle-1,-1\rangle$
For the vector $\frac{1}{2}\vec{v}$, $a_{1}=-1$ and $a_{2}=-1$:
$|\frac{1}{2}\vec{v}|=\sqrt{(-1)^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2}$
$|\vec{u}+\vec{v}|$
Add the vectors $\vec{u}$ and $\vec{v}$ together and simplify:
$\vec{u}+\vec{v}=\langle10,-1\rangle+\langle-2,-2\rangle=\langle8,-3\rangle$
For the vector $\vec{u}+\vec{v}$, $a_{1}=8$ and $a_{2}=-3$:
$|\vec{u}+\vec{v}|=\sqrt{8^{2}+(-3)^{2}}=\sqrt{64+9}=\sqrt{73}$
$|\vec{u}-\vec{v}|$
Subtract the vectors $\vec{u}$ and $\vec{v}$ and simplify:
$\vec{u}-\vec{v}=\langle10,-1\rangle-\langle-2,-2\rangle=\langle12,1\rangle$
For the vector $\vec{u}-\vec{v}$, $a_{1}=12$ and $a_{2}=1$:
$|\vec{u}-\vec{v}|=\sqrt{12^{2}+1^{2}}=\sqrt{144+1}=\sqrt{145}$
$|\vec{u}|-|\vec{v}|$
Subtract the magnitude of $\vec{v}$ from the magnitude of $\vec{u}$. These two magnitudes were found in the first two parts of this problem.
$|\vec{u}|=\sqrt{101}$ and $|\vec{v}|=2\sqrt{2}$
$|\vec{u}|-|\vec{v}|=\sqrt{101}-2\sqrt{2}$