Answer
(a) $\langle 19, -3 \rangle$
(b) $5\sqrt 2$
(c) $0$
(d) Yes.
Work Step by Step
We are given $\vec u=\langle 1,3 \rangle$ and $\vec u=\langle -6,2 \rangle$
(a) $\vec u-3\vec v=\langle 1-3(-6),3-3(2) \rangle=\langle 19, -3 \rangle$
(b) $|\vec u+\vec v|=|\langle 1-6,3+2 \rangle|=|\langle -5,5 \rangle|=\sqrt {(-5)^2+(5)^2}=5\sqrt 2$
(c) $\vec u\cdot\vec v=1(-6)+3(2)=0$
(d) Since $\vec u\cdot\vec v=0$, we conclude that $\vec u$ and $\vec v$ are penpendicular.
