Answer
$6$, $\langle 6,1,3 \rangle$, $\langle 2,-5,5 \rangle$, $\langle -1,-\frac{15}{2},5 \rangle$
Work Step by Step
Given $\vec u=\langle 4,-2,4 \rangle$ and $\vec u=\langle 2,3,-1 \rangle$, use the corresponding definitions and formulas.
1. $|\vec u|=\sqrt {(4)^2+(-2)^2+(4)^2}=6$
2. $\vec u+\vec v=\langle 4+2,-2+3,4-1 \rangle=\langle 6,1,3 \rangle$
3. $\vec u-\vec v=\langle 4-2,-2-3,4+1 \rangle=\langle 2,-5,5 \rangle$
4. $\frac{3}{4}\vec u-2\vec v=\langle \frac{3}{4}\times4-2\times2,\frac{3}{4}\times(-2)-2\times3,\frac{3}{4}\times4-2\times(-1) \rangle=\langle -1,-\frac{15}{2},3 \rangle$
