Answer
$\overrightarrow{a}+\overrightarrow{b} = 4\hat{i}+\hat{j}$
$4\overrightarrow{a}+ 2\overrightarrow{b}=18\hat{i} + 8\hat{j}$
$| \overrightarrow{a} |=\sqrt{34}$
$| \overrightarrow{a}-\overrightarrow{b} |=\sqrt{61}$
Work Step by Step
Given vectors $\overrightarrow{a}=5 \hat{i}+3\hat{j}$ and $\overrightarrow{b}=-\hat{i} -2 \hat{j}$
1) We find $\overrightarrow{a}+\overrightarrow{b}$ by adding the components of $\overrightarrow{a}$ and $\overrightarrow{b}$
$\overrightarrow{a}+\overrightarrow{b}
= (5-1)\hat{i}+(3-2)\hat{j}
= 4\hat{i}+\hat{j}$
2) We find $4\overrightarrow{a}+ 2\overrightarrow{b}$ by multiplying the components by their respective coefficients:
$4\overrightarrow{a} = 4(5)\hat{i} + 4(3)\hat{j}
=20\hat{i} + 12\hat{j}$
$2\overrightarrow{b} = 2(-1)\hat{i} + 2(-2)\hat{j}
=-2\hat{i} -4\hat{j}$
and then adding their components:
$4\overrightarrow{a}+ 2\overrightarrow{b}=
(20\hat{i} + 12\hat{j})+ (-2\hat{i} -4\hat{j})
=18\hat{i} + 8\hat{j}$
3)We find $| \overrightarrow{a} |$ with the distance formula:
$| \overrightarrow{a} |=\sqrt{(a_x)^2+(a_y)^2}$
$| \overrightarrow{a} |=\sqrt{(5)^2+(3)^2}=\sqrt{34}$
4) Finally, we find $| \overrightarrow{a}- \overrightarrow{b} |$ by first subtracting the components of $\overrightarrow{b}$ from $\overrightarrow{a}$
$\overrightarrow{a}-\overrightarrow{b} = (5-(-1))\hat{i} + (3-(-2))\hat{j}
=6\hat{i} +5)\hat{j}$
and then using the distance formula
$| \overrightarrow{a}-\overrightarrow{b} |=\sqrt{(6)^2+(5)^2}=\sqrt{36+25}=\sqrt{61}$
