Hey guys I am not sure if this is the right thread (can't seem to find one for vectors) but i am not sure if i'm on the right track here so i need some advice

Question is:

if $\displaystyle \vec{a}=3i-2j$ , $\displaystyle \vec{b}=-4i+4j$ , and $\displaystyle \vec{c}=6i+-9j$ express the following vectors in their simplest form:

(i)$\displaystyle -5\vec{b}$

(ii)$\displaystyle 2\vec{a}-\frac{1}{2}\vec{c}$

(iii)$\displaystyle \frac{2}{3}\vec{a}-\frac{1}{2}\vec{b}-\frac{1}{4}\vec{c}$

My Solution:

(i) $\displaystyle -5\left(-4i+4j \right)=20i+\left(-20j \right)$

(ii) $\displaystyle 2(3i-2j)-\frac{1}{2}(6i+(-9)j)$

$\displaystyle = 6i-4j-3i+(-\frac{9}{2})j$

$\displaystyle = (6-3=3i) -4j-\frac{9}{2}$

$\displaystyle = -\frac{4}{1}-\frac{9}{2}$

$\displaystyle =-\frac{8}{2}-\frac{9}{2}$

$\displaystyle =-\frac{17}{2}$

Answer? $\displaystyle 3i-\frac{17}{2}j$

(iii) $\displaystyle \frac{2}{3}(3i-2j)-\frac{1}{2}(-4i+4j)-\frac{1}{4}(6i+(-9)j)$

$\displaystyle (2i-\frac{4}{3}j)-(-2i+2j)-(\frac{3}{2}i+(-\frac{9}{4}j)$

$\displaystyle 2i+2i-\frac{3}{2}i$

$\displaystyle = \frac{4}{1}i+\frac{3}{2}i$

$\displaystyle = \frac{8}{2}+\frac{3}{2}$

$\displaystyle =\frac{11}{2}i$

$\displaystyle -\frac{4}{3}j+\frac{2}{1}j-\frac{9}{4}j$

$\displaystyle = -\frac{16}{12}j+\frac{24}{12}j-\frac{27}{12}j$

$\displaystyle = -\frac{19}{12}j$

Answer? $\displaystyle \frac{11}{2}i-\frac{19}{12}j$