91Ó°ÊÓ

Here, vector law of addition and subtraction is used to find the resultant of vector $\stackrel{→}{a}+\stackrel{→}{b}$and $\stackrel{→}{b}-\stackrel{→}{a}$in unit-vector notation. Further using the general formula for the magnitude and the angle, the magnitude and the angle of the given vector $\stackrel{→}{a}+\stackrel{→}{b}$and$\stackrel{→}{b}-\stackrel{→}{a}$ can be calculated.

Formulae to be used.

$$\begin{gathered} \stackrel{→}{a}+\stackrel{→}{b}=\left(a_x\right)\stackrel{Áåœ}{i}+\left(a_y\right)\stackrel{Áåœ}{j}+\left(b_x\right)\stackrel{Áåœ}{\mathrm{i}}+\left(b_y\right)\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{a}+\stackrel{→}{b}=\left(a_x+b_x\right)\stackrel{Áåœ}{i}+\left(a_y+b_y\right)\stackrel{Áåœ}{\mathrm{j}} \\ r=\backslash \stackrel{→}{a}+\stackrel{→}{b}\backslash =\sqrt{{\left(a_x+b_x\right)}^2+{\left(a_y+b_y\right)}^2} \\ \mathrm{θ}={\tan }^1\left(\frac{a_x+b_x}{a_y+b_y}\right) \end{gathered}$$

$$\begin{gathered} \stackrel{→}{b}-\stackrel{→}{a}=\left(b_x\right)\stackrel{Áåœ}{\mathrm{i}}+\left(b_y\right)\stackrel{Áåœ}{\mathrm{j}}-\left(a_x\right)\stackrel{Áåœ}{\mathrm{i}}-\left(a_y\right)\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{b}-\stackrel{→}{a}=\left(b_x-a_x\right)\stackrel{Áåœ}{i}+\left(b_y-a_y\right)\stackrel{Áåœ}{\mathrm{j}} \\ r=\backslash \stackrel{→}{b}-\stackrel{→}{a}\backslash =\sqrt{{\left(b_x-a_x\right)}^2+{\left(b_y-a_y\right)}^2} \\ \mathrm{θ}={\tan }^1\left(\frac{b_x-a_x}{b_y-a_y}\right) \\ \mathrm{Given}\mathrm{are}, \\ \stackrel{→}{a}=\left(3.00\mathrm{m}\right)\stackrel{Áåœ}{i}+\left(4.0m\right)\stackrel{Áåœ}{j} \\ \stackrel{→}{b}=\left(5.00m\right)\stackrel{Áåœ}{\mathrm{i}}+\left(-2.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

$$\begin{gathered} As\stackrel{→}{a}=\left(3.0m\right)\stackrel{Áåœ}{i}+\left(4.0m\right)\stackrel{Áåœ}{\mathrm{j}}and\stackrel{→}{b}=\left(5.0m\right)\stackrel{Áåœ}{\mathrm{i}}+\left(-2.0m\right)\stackrel{Áåœ}{\mathrm{j}} \\ \mathrm{so}, \\ \stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}=\left(3.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(4.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{j}}+\left(5.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(-2.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}=\left(8.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{i}}+\left(2.0\mathrm{m}\right)\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

$$\begin{gathered} \mathrm{Magnitude}\mathrm{of}\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}\mathrm{is} \\ \left|\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}\right|=\sqrt{{\left(8.0\mathrm{m}\right)}^2+{\left(2.0\mathrm{m}\right)}^2} \\ \left|\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}\right|=8.2\mathrm{m} \end{gathered}$$

The angle between $\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$and x axis is

${\tan }^{-1}\left(\frac{2.0\mathrm{m}}{8.0\mathrm{m}}\right)=14°$

$$\begin{gathered} \stackrel{→}{a}=\left(3.0m\right)\stackrel{Áåœ}{\mathrm{i}}+\left(4.0m\right)\stackrel{Áåœ}{\mathrm{j}}and\stackrel{→}{b}=\left(5.0m\right)\stackrel{Áåœ}{\mathrm{i}}+\left(-2.0m\right)\stackrel{Áåœ}{\mathrm{j}} \\ so, \\ \stackrel{→}{b}-\stackrel{→}{a}=\left(5.0m\right)\stackrel{Áåœ}{i}+\left(-2.0m\right)\stackrel{Áåœ}{j}-\left(3.0m\right)\stackrel{Áåœ}{i}+\left(4.0m\right)\stackrel{Áåœ}{j} \\ \stackrel{→}{b}-\stackrel{→}{a}=\left(2.0m\right)\stackrel{Áåœ}{i}-\left(6.0m\right)\stackrel{Áåœ}{j} \end{gathered}$$

$$\begin{gathered} \mathrm{Magnitude}\mathrm{of}\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}\mathrm{is} \\ \left|\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}\right|=\sqrt{{\left(2.0\mathrm{m}\right)}^2+{\left(-60\mathrm{m}\right)}^2} \\ \left|\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}\right|=6.3\mathrm{m} \end{gathered}$$

The angle between$$\begin{gathered} \stackrel{→}{b}-\stackrel{→}{a}\mathrm{and}\mathrm{x}\mathrm{axis}\mathrm{is} \\ {\tan }^{-1}\left(\frac{-6.0\mathrm{m}}{2.0}\right)=-72° \\ \mathrm{It}\mathrm{means}\mathrm{angle}\mathrm{is}180-72=108°\mathrm{relative}\mathrm{to}\stackrel{Áåœ}{\mathrm{i}}. \end{gathered}$$