Answer
$6\sqrt 2$, $\sqrt 5$, $12\sqrt 2$, $\frac{\sqrt 5}{2}$, $\sqrt {89}$, $\sqrt {65}$, $6\sqrt 2-\sqrt 5$
Work Step by Step
Given $\vec u=\langle -6, 6 \rangle$ and $\vec v=\langle -2, -1 \rangle$, use for magnitude formula, we have:
(i) $|\vec u|=|\langle -6, 6 \rangle|=\sqrt {(-6)^2+6^2}=6\sqrt 2$
(ii) $|\vec v|=|\langle -2, -1 \rangle|=\sqrt {(-2)^2+(-1)^2}=\sqrt 5$
(iii) $|2\vec u|=|\langle -12, 12 \rangle|=\sqrt {(-12)^2+12^2}=12\sqrt 2$
(iv) $|\frac{1}{2}\vec v|=|\langle -1, -1/2 \rangle|=\sqrt {(-1)^2+(-\frac{1}{2})^2}=\frac{\sqrt 5}{2}$
(v) $|\vec u+\vec v|=|\langle -6-2, 6-1 \rangle|=\sqrt {(-8)^2+(5)^2}=\sqrt {89}$
(vi) $|\vec u-\vec v|=|\langle -6+2, 6+1 \rangle|=\sqrt {(-4)^2+(7)^2}=\sqrt {65}$
(vii) Use results from (i) and (ii), $|\vec u|-|\vec v|=6\sqrt 2-\sqrt 5$