Answer
$\overrightarrow{a}+\overrightarrow{b} =6\hat{i}-3\hat{j} -2\hat{k}$
$4\overrightarrow{a}+ 2\overrightarrow{b}= 20\hat{i} -12\hat{j}+0\hat{k}$
$| \overrightarrow{a} |=\sqrt{29}$
$| \overrightarrow{a}-\overrightarrow{b} |=7$
Work Step by Step
Given vectors $\overrightarrow{a}=4\hat{i} -3\hat{j}+ 2\hat{k}$ and $\overrightarrow{b}=2\hat{i} + 0\hat{j}-4 \hat{k}$
1) We find $\overrightarrow{a}+\overrightarrow{b}$ by adding the components of $\overrightarrow{a}$ and $\overrightarrow{b}$
$\overrightarrow{a}+\overrightarrow{b}
= (4+2)\hat{i}+(-3+0)\hat{j}+(2+(-4))\hat{k}
= 6\hat{i}-3\hat{j} -2\hat{k}$
2) We find $4\overrightarrow{a}+ 2\overrightarrow{b}$ by multiplying the components by their respective coefficients:
$4\overrightarrow{a} = 4(4)\hat{i} + 4(-3)\hat{j}+4(2)\hat{k}
=16\hat{i} -12\hat{j}+8\hat{k}$
$2\overrightarrow{b} = 2(2)\hat{i} + 2(0)\hat{j}+2(-4)\hat{k}
=4\hat{i} +0\hat{j} -8\hat{k}$
and then adding their components:
$4\overrightarrow{a}+ 2\overrightarrow{b}=
(16+4)\hat{i} +(-12+0)\hat{j}+(8-8)\hat{k}
=20\hat{i} -12\hat{j}+0\hat{k}$
3)We find $| \overrightarrow{a} |$ with the distance formula:
$| \overrightarrow{a} |=\sqrt{(a_x)^2+(a_y)^2+(a_z)^2}$
$| \overrightarrow{a} |=\sqrt{(4)^2+(-3)^2+(2)^2}=\sqrt{29}$
4) Finally, we find $| \overrightarrow{a}- \overrightarrow{b} |$ by first subtracting the components of $\overrightarrow{b}$ from $\overrightarrow{a}$
$\overrightarrow{a}-\overrightarrow{b} = (4-(2))\hat{i} + (-3-(0))\hat{j}+(2-(-4))\hat{k}$
$\overrightarrow{a}-\overrightarrow{b}=2\hat{i} -3\hat{j}+6\hat{k}$
and then using the distance formula
$| \overrightarrow{a}-\overrightarrow{b} |=
\sqrt{(2)^2+(-3)^2+(6)^2}=\sqrt{49}=7$
