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Expert-verified Found in: Page 57 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Three vectors $\stackrel{\mathbf{\to }}{\mathbf{a}}{\mathbf{,}}\stackrel{\mathbf{\to }}{\mathbf{b}}{\mathbf{}}{\mathbit{a}}{\mathbit{n}}{\mathbit{d}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{c}}$ each have a magnitude of 50m and lie in an xy plane. Their directions relative to the positive direction of the x axis are ${\mathbf{30}}{\mathbf{°}}$,${\mathbf{195}}{\mathbf{°}}$, and ${\mathbf{315}}{\mathbf{°}}$, respectively. What are (a) the magnitude and (b) the angle of the vector localid="1656259759790" $\stackrel{\mathbf{\to }}{\mathbf{a}}+\stackrel{\mathbf{\to }}{\mathbf{b}}+\stackrel{\mathbf{\to }}{\mathbf{c}}$ , and (c) the magnitude and (d) the angle of $\stackrel{\mathbf{\to }}{\mathbf{a}}+\stackrel{\mathbf{\to }}{\mathbf{b}}{\mathbf{+}}\stackrel{\mathbf{\to }}{\mathbf{c}}$ ? What are the (e) magnitude and (f) angle of a fourth vector $\stackrel{\mathbf{\to }}{\mathbf{d}}$ such that $\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right){\mathbit{-}}{\mathbit{}}\left(\stackrel{\to }{c}+\stackrel{\to }{d}\right){\mathbit{=}}{\mathbit{0}}$ ?

Magnitude of vector $\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}$ is

Angle of vector $\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}$ is

Magnitude of vector $\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}$ is

Angle of vector $\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}$ is

Magnitude of fourth vector is

Angle of fourth vector is

See the step by step solution

## Step 1: To understand the concept

Here, vector law of addition and subtraction is used to find the resultant of the given vector. Further using the general formula for the magnitude and the angle, the magnitude and the angle of the given vector can be calculated.

Formulae

$\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}{=}\left({a}_{x}\right)\stackrel{⏜}{\mathrm{i}}{+}\left({a}_{y}\right)\stackrel{⏜}{j}{+}\left({b}_{x}\right)\stackrel{⏜}{\mathrm{i}}{+}\left({b}_{y}\right)\stackrel{⏜}{j}{+}\left({c}_{x}\right)\stackrel{⏜}{\mathrm{i}}{+}\left({c}_{y}\right)\stackrel{⏜}{j}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}{=}\left({a}_{x}+{b}_{x}+{c}_{x}\right)\stackrel{⏜}{\mathrm{i}}{+}\left({a}_{y}+{b}_{y}+{c}_{y}\right)\stackrel{⏜}{j}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{=}\left({a}_{x}+{b}_{x}\right)\stackrel{⏜}{\mathrm{i}}{+}\left({a}_{y}+{b}_{y}\right)\stackrel{⏜}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}{r}{=}\left|\stackrel{\mathit{\to }}{\mathit{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right|{=}\sqrt{{\left({a}_{x}+{b}_{x}+{c}_{x}\right)}^{2}+{\left({a}_{y}+{b}_{y}+{c}_{y}\right)}^{2}}\phantom{\rule{0ex}{0ex}}{\theta }{=}{{\mathrm{tan}}}^{-1}\left(\frac{{a}_{x}+{b}_{x}+{c}_{x}}{{a}_{y}+{b}_{y}+{c}_{y}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{Given}}{}{\mathrm{are}}\phantom{\rule{0ex}{0ex}}\stackrel{\mathit{\to }}{\mathit{a}}{=}\left(50\mathrm{m}\right){\mathrm{cos}}\left(30\right)\stackrel{⏜}{\mathrm{i}}{+}\left(50\mathrm{m}\right){\mathrm{sin}}\left(30\right)\stackrel{⏜}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{b}}{=}\left(50\mathrm{m}\right){\mathrm{cos}}\left(195\right)\stackrel{⏜}{\mathrm{i}}{+}\left(50\mathrm{m}\right){\mathrm{sin}}\left(195\right)\stackrel{⏜}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{c}}{=}\left(50\mathrm{m}\right){\mathrm{cos}}\left(315\right)\stackrel{⏜}{\mathrm{i}}{+}\left(50\mathrm{m}\right){\mathrm{sin}}\left(315\right)\stackrel{⏜}{\mathrm{j}}$

## Step 2: To find magnitude of vector a→+b→+c→

Using the above values the vector $\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}$ can be written as

$\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}{=}\left(30.4\right)\stackrel{⏜}{i}\left(-23.3m\right)\stackrel{⏜}{j}\phantom{\rule{0ex}{0ex}}{\mathrm{Magnitude}}{}{\mathrm{of}}{}\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}{}{\mathrm{is}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right|{=}\sqrt{\left(30.4{\mathrm{m}}^{2}\right)+{\left(-23.3\mathrm{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right|{=}{38}{\mathrm{m}}$

## Step 3: To find the angle between vector a→+b→+c→ and x axis

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

${\mathrm{tan}}^{-1}\left(\frac{-23.3\mathrm{m}}{30.4\mathrm{m}}\right)=-37.5°$

This is equivalent to $37.5°$clockwise from the +x axis and $322.5°$counterclockwise from +x axis

## Step 4: To find magnitude of vector a→+b→+c→

$\stackrel{\to }{\mathrm{a}}{+}\stackrel{\to }{\mathrm{b}}{+}\stackrel{\to }{\mathrm{c}}{=}\left(127\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}{+}\left(2.60\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right|{=}\sqrt{{\left(127\mathrm{m}\right)}^{2}+{\left(2.60\mathrm{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}\right|{=}{1}{.}{30}{×}{{10}}^{{2}}{\mathrm{m}}$

## Step 5: To find the angle between vector a→+b→+c→ and x axis

The angle between$\stackrel{\mathit{\to }}{\mathit{a}}{\mathit{+}}\stackrel{\mathit{\to }}{\mathit{b}}{\mathit{+}}\stackrel{\mathit{\to }}{\mathit{c}}$ and x axis is

${\mathrm{tan}}^{-1}\left(\frac{26.5\mathrm{m}}{127\mathrm{m}}\right)=1.2°$

Therefore, the angle between $\stackrel{\mathit{\to }}{a}{\mathit{+}}\stackrel{\mathit{\to }}{b}{\mathit{+}}\stackrel{\mathit{\to }}{c}$ and +x axis is $1.2°$

## Step 6: To find magnitude of fourth vector d→

$\stackrel{\to }{d}=\stackrel{\mathit{\to }}{a}{\mathit{+}}\stackrel{\mathit{\to }}{b}{\mathit{+}}\stackrel{\mathit{\to }}{c}{\mathit{=}}\left(-40.4\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}{+}\left(47.4\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{d}}\right|{=}\sqrt{{\left(-40.4\mathrm{m}\right)}^{2}+{\left(47.4\mathrm{m}\right)}^{2}}{=}{62}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{\mathrm{d}}\right|{=}{62}{\mathrm{m}}$

## Step 7: To find the angle between vector d→ and x axis

The angle between $\stackrel{\to }{d}$ and +x axis is

${\mathrm{tan}}^{-1}\left(\frac{47.4\mathrm{m}}{-40.4\mathrm{m}}\right)=50°$

As vector $\stackrel{\to }{d}$ is in third quadrant so,

$180°-50°=130°$

Therefore, the angle between $\stackrel{\to }{d}$ and +x axis is ### Want to see more solutions like these? 