Answer
$|\vec{u}|=\sqrt{5}$
$|\vec{v}|=\sqrt{13}$
$|2\vec{u}|=2\sqrt{5}$
$|\frac{1}{2}\vec{v}|=\dfrac{\sqrt{13}}{2}$
$|\vec{u}+\vec{v}|=\sqrt{26}$
$|\vec{u}-\vec{v}|=\sqrt{10}$
$|\vec{u}|-|\vec{v}|=\sqrt{5}-\sqrt{13}$
Work Step by Step
$\vec{u}=2\vec{i}+\vec{j},$ $\vec{v}=3\vec{i}-2\vec{j}$
To find the magnitud of a vector $\vec{v}=\langle a_{1},a_{2}\rangle$, just use the formula $|\vec{v}|=\sqrt{a_{1}^{2}+a_{2}^{2}}$
$|\vec{u}|$
For the vector $\vec{u}$, $a_{1}=2$ and $a_{2}=1$
$|\vec{u}|=\sqrt{2^{2}+1^{2}}=\sqrt{4+1}=\sqrt{5}$
$|\vec{v}|$
For the vector $\vec{v}$, $a_{1}=3$ and $a_{2}=-2$
$|\vec{v}|=\sqrt{3^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}$
$|2\vec{u}|$
Multiply both components of $\vec{u}$ by $2$:
$2\vec{u}=2(2\vec{i}+\vec{j})=4\vec{i}+2\vec{j}$
For the vector $2\vec{u}$, $a_{1}=4$ and $a_{2}=2$:
$|2\vec{u}|=\sqrt{4^{2}+2^{2}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
$|\frac{1}{2}\vec{v}|$
Multiply both components of $\vec{v}$ by $\frac{1}{2}$:
$\frac{1}{2}\vec{v}=\dfrac{1}{2}(3\vec{i}-2\vec{j})=\dfrac{3}{2}\vec{i}-\vec{j}$
For the vector $\frac{1}{2}\vec{v}$, $a_{1}=\frac{3}{2}$ and $a_{2}=-1$:
$|\frac{1}{2}\vec{v}|=\sqrt{\Big(\dfrac{3}{2}\Big)^{2}+(-1)^{2}}=\sqrt{\dfrac{9}{4}+1}=\sqrt{\dfrac{13}{4}}=\dfrac{\sqrt{13}}{2}$
$|\vec{u}+\vec{v}|$
Add the vectors $\vec{u}$ and $\vec{v}$ together and simplify:
$\vec{u}+\vec{v}=(2\vec{i}+\vec{j})+(3\vec{i}-2\vec{j})=(2+3)\vec{i}+(1-2)\vec{j}=...$
$...=5\vec{i}-\vec{j}$
For the vector $\vec{u}+\vec{v}$, $a_{1}=5$ and $a_{2}=-1$:
$|\vec{u}+\vec{v}|=\sqrt{5^{2}+(-1)^{2}}=\sqrt{25+1}=\sqrt{26}$
$|\vec{u}-\vec{v}|$
Subtract the vectors $\vec{u}$ and $\vec{v}$ and simplify:
$\vec{u}-\vec{v}=(2\vec{i}+\vec{j})-(3\vec{i}-2\vec{j})=(2-3)\vec{i}+(1+2)\vec{j}=...$
$...=-\vec{i}+3\vec{j}$
For the vector $\vec{u}-\vec{v}$, $a_{1}=-1$ and $a_{2}=3$:
$|\vec{u}-\vec{v}|=\sqrt{(-1)^{2}+3^{2}}=\sqrt{1+9}=\sqrt{10}$
$|\vec{u}|-|\vec{v}|$
We need to subtract the magnitude of $\vec{v}$ from the magnitude of $\vec{u}$. The magnitudes of $\vec{u}$ and $\vec{v}$ were found in the first two parts of this problem and they are $\sqrt{5}$ and $\sqrt{13}$, respectively:
$|\vec{u}|-|\vec{v}|=\sqrt{5}-\sqrt{13}$
