Question

Let $\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$Find the magnitudes of

a.$\overrightarrow{E}and\overrightarrow{F}$b.$\overrightarrow{E}+\overrightarrow{F}$c.$-\overrightarrow{E}-2\overrightarrow{F}$

Short answer

(a) The magnitude of $\overrightarrow{E}$is $\sqrt{13}$and $\overrightarrow{F}$is $\sqrt{8}.$

(b) The magnitude of $\overrightarrow{E}+\overrightarrow{F}$is $\sqrt{17}.$

(c) The magnitude of$-\overrightarrow{E}+2\overrightarrow{F}$is$\sqrt{37}.$

Step-by-step solution

Part (a) Step 1. Given information

It is given that $\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$We need to determine the magnitude of $E$and$F.$

Step 2. Calculation

$\overrightarrow{E}=2\hat{i}+3\hat{j}.$

$E_x=2and{E}_y=3.$

$\left|\overrightarrow{E}\right|=\sqrt{{\left(E_x\right)}^2+{\left(E_y\right)}^2}.$

$=\sqrt{{\left(2\right)}^2+{\left(3\right)}^2}.$

$=\sqrt{4+9}.$

$=\sqrt{13}.$

$\overrightarrow{F}=2\hat{i}-2\hat{j}.$

$F_x=2,F_y=-2.$

$\left|\overrightarrow{F}\right|=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2}.$

$=\sqrt{{\left(2\right)}^2+{\left(-2\right)}^2}.$

$=\sqrt{4+4}.$

$=\sqrt{8}.$

Part (b) Step 1. Given information

It is given that,$\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$We need to determine the magnitude of$\overrightarrow{E}+\overrightarrow{F}.$

Step 2. Calculation

$\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$

$\overrightarrow{E}+\overrightarrow{F}=\left(2\hat{i}+3\hat{j}\right)+\left(2\hat{i}-2\hat{j}.\right)$

$=4\hat{i}-\hat{j}.$

role="math" localid="1649179077562" ${\left(E+F\right)}_x=4,{\left(E+F\right)}_y=-1.$

$\left|\overrightarrow{E}+\overrightarrow{F}\right|=\sqrt{{\left(E+F\right)}_x^2+{\left(E+F\right)}_y^2}.$

$=\sqrt{{\left(4\right)}^2+{\left(-1\right)}^2.}$

$=\sqrt{16+1}.$

$=\sqrt{17}.$

Part (c) Step 1. Given information

It is given that, $\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$We need to determine the magnitude of$-\overrightarrow{E}+2\overrightarrow{F}.$

Step 2. Calculation

$\overrightarrow{E}=2\hat{i}+3\hat{j}and\overrightarrow{F}=2\hat{i}-2\hat{j}.$

$-\overrightarrow{E}-2\overrightarrow{F}.$

$=-\left(2\hat{i}+3\hat{j}\right)-2\left(2\hat{i}-2\hat{j}\right).$

$=-2\hat{i}-3\hat{j}-4\hat{i}+4\hat{j}.$

$=-6\hat{i}+\hat{j}.$

${\left(-E+2F\right)}_x=-6,{\left(-E+2F\right)}_y=1.$

$\left|-\overrightarrow{E}+2\overrightarrow{F}\right|=\sqrt{{{\left(-E+2F\right)}^2}_x+{{\left(-E+2F\right)}^2}_y}.$

$=\sqrt{{\left(-6\right)}^2+{\left(1\right)}^2.}$

$=\sqrt{36+1}.$

$=\sqrt{37}.$