Answer
$\overrightarrow{a}+\overrightarrow{b}= <13,-1,-3>$
$4\overrightarrow{a}+ 2\overrightarrow{b}= <42,0,-14>$
$| \overrightarrow{a} |=9$
$| \overrightarrow{a}-\overrightarrow{b} |=\sqrt{43}$
Work Step by Step
Given vectors $\overrightarrow{a}=<8,1,-4>$ and $\overrightarrow{b}=<5,-2,1>$
1) We find $\overrightarrow{a}+\overrightarrow{b}$ by adding the components of $\overrightarrow{a}$ and $\overrightarrow{b}$
$\overrightarrow{a}+\overrightarrow{b}
= <(8+5),(1+(-2)),(-4+1)> $
$\overrightarrow{a}+\overrightarrow{b}= <13,-1,-3>$
2) We find $4\overrightarrow{a}+ 2\overrightarrow{b}$ by multiplying the components by their respective coefficients:
$4\overrightarrow{a} = <4(8),4(1),4(-4)>=<32,4,-16>$
$2\overrightarrow{b} = <2(5),2(-2),2(1)>=<10,-4,2>$
and then adding their components:
$4\overrightarrow{a}+ 2\overrightarrow{b}=<42,0,-14>$
3) We find $| \overrightarrow{a} |$ with the distance formula:
$| \overrightarrow{a} |=\sqrt{(a_x)^2+(a_y)^2+(a_z)^2}$
$| \overrightarrow{a} |=\sqrt{(8)^2+(1)^2+(-4)^2}=\sqrt{81}=9$
4) Finally, we find $| \overrightarrow{a}- \overrightarrow{b} |$ by first subtracting the components of $\overrightarrow{b}$ from $\overrightarrow{a}$
$\overrightarrow{a}-\overrightarrow{b} = <(8-5),(1-(-2)),(-4-1)>$
$\overrightarrow{a}-\overrightarrow{b}=<3,-3,-5>$
and then using the distance formula
$| \overrightarrow{a}-\overrightarrow{b} |=
\sqrt{(3)^2+(-3)^2+(5)^2}=\sqrt{43}$
