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Expert-verified Found in: Page 60 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Here are three displacements, each measured in meters: ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{i}}{\mathbf{+}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{j}}{\mathbf{-}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{k}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{i}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{j}}{\mathbf{+}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{k}}$ ,and${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{i}}{\mathbf{+}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{j}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{⏜}}{\mathbf{k}}$. (a) What is $\stackrel{\mathbf{\to }}{\mathbf{r}}{\mathbf{=}}{\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{1}}}{\mathbf{+}}{\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{2}}}{\mathbf{+}}{\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{3}}}$ ? (b) What is the angle between$\stackrel{\mathbf{\to }}{\mathbf{r}}$ and the positive z axis? (c) What is the component of ${{\mathbit{d}}}_{{\mathbf{1}}}$ along the direction of ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{2}}}$ (d) What is the component of ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{1}}}$ that is perpendicular to the direction of $\stackrel{\mathbf{\to }}{{\mathbf{d}}_{\mathbf{2}}}$ and in the plane of role="math" localid="1658465314757" ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{1}}}$and ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{2}}}$ (Hint: For (c), consider Eq 3-20. and Fig, 3-18; for (d), consider Eq.3-24.)

a) The displacement $\stackrel{\to }{r}$ is $9.0\stackrel{⏜}{i}+6.0\stackrel{⏜}{j}-7.0\stackrel{⏜}{k} m$

b) The angle between$\stackrel{\to }{r}$and the positive z-axis is $\theta =123°$

c) The component of $\stackrel{\to }{{d}_{1}}$ along the direction of role="math" localid="1658465562638" ${\stackrel{\to }{d}}_{2}is=-3.2\mathrm{m}$.

d) The component of ${\stackrel{\to }{d}}_{2}$ in the perpendicular direction of ${\stackrel{\to }{d}}_{2}\mathrm{is}8.2\mathrm{m}$.

See the step by step solution

## Step 1: Given data

The given vectors are,

${\stackrel{\to }{\mathrm{d}}}_{{1}}{=}{4}{.}{0}\stackrel{⏜}{\mathrm{i}}{+}{5}{.}{0}\stackrel{⏜}{\mathrm{j}}{-}{6}{.}{0}\stackrel{⏜}{\mathrm{k}}$

${\stackrel{\to }{\mathrm{d}}}_{{2}}{=}{-}{1}{.}{0}\stackrel{⏜}{\mathrm{i}}{+}{2}{.}{0}\stackrel{⏜}{\mathrm{j}}{+}{3}{.}{0}\stackrel{⏜}{\mathrm{k}}$

${\stackrel{\to }{\mathrm{d}}}_{{3}}{=}{4}{.}{0}\stackrel{⏜}{\mathrm{i}}{+}{3}{.}{0}\stackrel{⏜}{\mathrm{j}}{+}{2}{.}{0}\stackrel{⏜}{\mathrm{k}}$

## Step 2: Understanding the concept

Vector addition and subtraction can be used to find the displacement and magnitude can be found using the formula. To find the angle between $\stackrel{\mathbf{\to }}{\mathbf{r}}$ and positive z axis we can use a scalar product of two vectors. Unit vector along z axis is. The components of ${\stackrel{\mathbf{\to }}{\mathbf{d}}}_{{\mathbf{1}}}$can be found by using the property of the scalar product of two vectors.

Formula:

$\stackrel{\to }{\mathrm{r}}{=}{\stackrel{\to }{\mathrm{d}}}_{{1}}{+}{\stackrel{\to }{\mathrm{d}}}_{{2}}{+}{\stackrel{\to }{\mathrm{d}}}_{{3}}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}{.}\left(\mathrm{i}\right)$

${\stackrel{\to }{\mathrm{d}}}_{1}.{\stackrel{\to }{\mathrm{d}}}_{1}={\mathrm{d}}_{1}{\mathrm{d}}_{2}\mathrm{cos\phi }.............\left(\mathrm{ii}\right)$

## Step 3: (a) Calculate the displacement r→

From equation (i), we can calculate the displacement vector $\stackrel{\to }{r}$ as follows.

$\stackrel{\to }{r}=\left(4.0\stackrel{⏜}{i}+5.0\stackrel{⏜}{j}-6.0\stackrel{⏜}{k}\right)-\left(-1.0\stackrel{⏜}{i}+2.0\stackrel{⏜}{j}+3.0\stackrel{⏜}{k}\right)+\left(4.0\stackrel{⏜}{i}+3.0\stackrel{⏜}{j}-2.0\stackrel{⏜}{k}\right)\phantom{\rule{0ex}{0ex}}=4.0\stackrel{⏜}{i}+5.0\stackrel{⏜}{j}-6.0\stackrel{⏜}{k}+1.0\stackrel{⏜}{i}-2.0\stackrel{⏜}{j}-3.0\stackrel{⏜}{k}+4.0\stackrel{⏜}{i}+3.0\stackrel{⏜}{j}+2.0\stackrel{⏜}{k}\phantom{\rule{0ex}{0ex}}=9.0\stackrel{⏜}{i}+6.0\stackrel{⏜}{j}-7.0\stackrel{⏜}{k}$

Therefore, the displacement$\stackrel{\to }{r}$ is$9.0\stackrel{⏜}{i}+6.0\stackrel{⏜}{j}-7.0\stackrel{⏜}{k}$.

## Step 4: (b) Calculate the angle betweenr→and positive z axis

The magnitude of r is calculated as,

$r=\sqrt{{\left(9.0\right)}^{2}+{\left(6.0\right)}^{2}+{\left(-7.0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=12.9\mathrm{m}$

The angle between$\stackrel{\to }{r}$and positive z axis is θ

localid="1658466997778" $\mathrm{cos}\theta =\frac{\stackrel{\to }{r}.\stackrel{⏜}{k}}{\left|\stackrel{\to }{r}\right|}\phantom{\rule{0ex}{0ex}}=\frac{-7.0}{12.0}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}\left(-0.543\right)\phantom{\rule{0ex}{0ex}}=123°$

The angle betweenlocalid="1658467048657" $\stackrel{\to }{r}$and positive z axis is $123°$.

## Step 5: (c) Calculate the component of d→1 along the direction of d→2

The component of ${\stackrel{\to }{d}}_{1}$along the direction of ${\stackrel{\to }{d}}_{1}={\stackrel{\to }{d}}_{||}$

${\stackrel{\to }{d}}_{▯}={\stackrel{\to }{d}}_{1}.\stackrel{⏜}{u}\phantom{\rule{0ex}{0ex}}={d}_{1}\mathrm{cos}\varphi .............\left(\mathrm{iii}\right)$

$\stackrel{⏜}{u}$ is a unit vector

${\stackrel{\to }{d}}_{1}.{\stackrel{\to }{d}}_{2}={d}_{1}{d}_{2}\mathrm{cos}\varphi \phantom{\rule{0ex}{0ex}}\mathrm{cos}\varphi =\frac{{\stackrel{\to }{d}}_{1}\stackrel{\to }{{d}_{2}}}{{d}_{1}{d}_{2}}.............\left(iv\right)$

Equation (iii) becomes as

$\stackrel{\to }{{d}_{▯}}={d}_{1}\mathrm{cos}\varphi \phantom{\rule{0ex}{0ex}}={d}_{1}\left(\frac{{\stackrel{\to }{d}}_{1}.{\stackrel{\to }{d}}_{2}}{{d}_{1}{d}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\stackrel{\to }{d}}_{1}.{\stackrel{\to }{d}}_{2}}{{d}_{2}}$

Now, calculate the dot product ${\stackrel{\to }{d}}_{1}.{\stackrel{\to }{d}}_{2}$ as,

role="math" localid="1658467900456" ${\stackrel{\to }{d}}_{1}.{\stackrel{\to }{d}}_{2}={d}_{1x}{d}_{2x}+{d}_{1y}{d}_{2y}+{d}_{1z}+{d}_{2z}\phantom{\rule{0ex}{0ex}}=\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right)$

Substitute it in the above equation.

role="math" localid="1658468009605" $\stackrel{\to }{{d}_{▯}}=\frac{\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right)}{\sqrt{{\left(-1.0\right)}^{2}\left(2.0\right)+{\left(3.0\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=-3.2\mathrm{m}$

## Step 6: (d) Calculate the component of d→1 in the perpendicular direction of d→2

The component of ${\stackrel{\to }{d}}_{1}$in the perpendicular direction of $\stackrel{\to }{{d}_{2}}=\stackrel{\to }{{d}_{\perp }}$

${\stackrel{\to }{d}}_{1}={\stackrel{\to }{d}}_{\perp }+{\stackrel{\to }{d}}_{▯}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{d}}_{\perp }={\stackrel{\to }{d}}_{1}-{\stackrel{\to }{d}}_{▯}\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{d}}_{\perp }^{2}={\stackrel{\to }{d}}_{1}^{2}-{d}_{▯}\phantom{\rule{0ex}{0ex}}{d}_{\perp }=\sqrt{\left[{\left(4.0\right)}^{2}+{\left(5.0\right)}^{2}+{\left(-6.0\right)}^{2}-{\left(-3.2\right)}^{2}\right]}\phantom{\rule{0ex}{0ex}}=8.2m$

Therefore, the component of ${\stackrel{\to }{d}}_{1}$ in the perpendicular direction of ${\stackrel{\to }{d}}_{2}$is 8.2 m ### Want to see more solutions like these? 