Question

For the displacement vectors $\overrightarrow{\mathbf{a}}\mathbf{=}\left(3.0m\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(4.0m\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}$and $\overrightarrow{\mathbf{b}}\mathbf{=}\left(5.0m\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(-2.0m\right)\stackrel{\mathbf{⏜}}{\mathbf{j}}$, give $\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}$in (a) unit-vector notation, and as (b) a magnitude and (c) an angle (relative to $\stackrel{\mathbf{⏜}}{\mathbf{i}}$). Now give $\overrightarrow{\mathbf{b}}\mathbf{-}\overrightarrow{\mathbf{a}}$in (d) unit-vector notation, and as (e) a magnitude and (f) an angle.

Short answer

$\overrightarrow{\mathrm{a}}+\overrightarrow{b}\mathrm{in}\mathrm{unit}-\mathrm{vector}\mathrm{notation}\mathrm{is}\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\left(8.0\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(2.0\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}$

Step-by-step solution

To understand the concept of the given problem

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

Formulae to be used.

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

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

To write in unit-vector notation

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

To calculate magnitude of a→+b→

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

To calculate angle between  a→+b→ and x axis

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

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

To write b→-a→ in unit-vector notation

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

To find magnitude of b→-a→

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

To find angle between b→-a→ and x axis

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