Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Here are three displacements, each measured in meters: ${\vec{d}}_{1} = 4.0 \overset{⏜}{i} + 5.0 \overset{⏜}{j} - 6.0 \overset{⏜}{k}, {\vec{d}}_{2} = - 1.0 \overset{⏜}{i} + 2.0 \overset{⏜}{j} + 3.0 \overset{⏜}{k}$ ,and ${\vec{d}}_{3} = 4.0 \overset{⏜}{i} + 3.0 \overset{⏜}{j} + 2.0 \overset{⏜}{k}$ . (a) What is $\vec{r} = {\vec{d}}_{1} + {\vec{d}}_{2} + {\vec{d}}_{3}$ ? (b) What is the angle between $\vec{r}$ and the positive z axis? (c) What is the component of $d_{1}$ along the direction of ${\vec{d}}_{2}$ (d) What is the component of ${\vec{d}}_{1}$ that is perpendicular to the direction of $\vec{d_{2}}$ and in the plane of role="math" localid="1658465314757" ${\vec{d}}_{1}$ and ${\vec{d}}_{2}$ (Hint: For (c), consider Eq 3-20. and Fig, 3-18; for (d), consider Eq.3-24.)

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

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

c) The component of $\vec{d_{1}}$ along the direction of role="math" localid="1658465562638" ${\vec{d}}_{2} i s = - 3.2 m$ .

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given vectors are,

${\vec{d}}_{1} = 4.0 \overset{⏜}{i} + 5.0 \overset{⏜}{j} - 6.0 \overset{⏜}{k}$

${\vec{d}}_{2} = - 1.0 \overset{⏜}{i} + 2.0 \overset{⏜}{j} + 3.0 \overset{⏜}{k}$

${\vec{d}}_{3} = 4.0 \overset{⏜}{i} + 3.0 \overset{⏜}{j} + 2.0 \overset{⏜}{k}$

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 $\vec{r}$ and positive z axis we can use a scalar product of two vectors. Unit vector along z axis is. The components of ${\vec{d}}_{1}$ can be found by using the property of the scalar product of two vectors.

Formula:

$\vec{r} = {\vec{d}}_{1} + {\vec{d}}_{2} + {\vec{d}}_{3} . . . . . . . . . . . . (i)$

${\vec{d}}_{1} . {\vec{d}}_{1} = d_{1} d_{2} cosφ . . . . . . . . . . . . . (ii)$

Step 3: (a) Calculate the displacement r→

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

$\vec{r} = (4.0 \overset{⏜}{i} + 5.0 \overset{⏜}{j} - 6.0 \overset{⏜}{k}) - (- 1.0 \overset{⏜}{i} + 2.0 \overset{⏜}{j} + 3.0 \overset{⏜}{k}) + (4.0 \overset{⏜}{i} + 3.0 \overset{⏜}{j} - 2.0 \overset{⏜}{k}) = 4.0 \overset{⏜}{i} + 5.0 \overset{⏜}{j} - 6.0 \overset{⏜}{k} + 1.0 \overset{⏜}{i} - 2.0 \overset{⏜}{j} - 3.0 \overset{⏜}{k} + 4.0 \overset{⏜}{i} + 3.0 \overset{⏜}{j} + 2.0 \overset{⏜}{k} = 9.0 \overset{⏜}{i} + 6.0 \overset{⏜}{j} - 7.0 \overset{⏜}{k}$

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

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

The magnitude of r is calculated as,

$r = \sqrt{{(9.0)}^{2} + {(6.0)}^{2} + {(- 7.0)}^{2}} = 12.9 m$

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

localid="1658466997778" $\cos θ = \frac{\vec{r} . \overset{⏜}{k}}{|\vec{r}|} = \frac{- 7.0}{12.0} θ = \cos^{- 1} (- 0.543) = 123 °$

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

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

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

${\vec{d}}_{▯} = {\vec{d}}_{1} . \overset{⏜}{u} = d_{1} \cos ϕ . . . . . . . . . . . . . (iii)$

$\overset{⏜}{u}$ is a unit vector

${\vec{d}}_{1} . {\vec{d}}_{2} = d_{1} d_{2} \cos ϕ \cos ϕ = \frac{{\vec{d}}_{1} \vec{d_{2}}}{d_{1} d_{2}} . . . . . . . . . . . . . (i v)$

Equation (iii) becomes as

$\vec{d_{▯}} = d_{1} \cos ϕ = d_{1} (\frac{{\vec{d}}_{1} . {\vec{d}}_{2}}{d_{1} d_{2}}) = \frac{{\vec{d}}_{1} . {\vec{d}}_{2}}{d_{2}}$

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

role="math" localid="1658467900456" ${\vec{d}}_{1} . {\vec{d}}_{2} = d_{1 x} d_{2 x} + d_{1 y} d_{2 y} + d_{1 z} + d_{2 z} = (4.0) (- 1.0) + (5.0) (2.0) + (- 6.0) (3.0)$

Substitute it in the above equation.

role="math" localid="1658468009605" $\vec{d_{▯}} = \frac{(4.0) (- 1.0) + (5.0) (2.0) + (- 6.0) (3.0)}{\sqrt{{(- 1.0)}^{2} (2.0) + {(3.0)}^{2}}} = - 3.2 m$

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

The component of ${\vec{d}}_{1}$ in the perpendicular direction of $\vec{d_{2}} = \vec{d_{⊥}}$

${\vec{d}}_{1} = {\vec{d}}_{⊥} + {\vec{d}}_{▯} {\vec{d}}_{⊥} = {\vec{d}}_{1} - {\vec{d}}_{▯} {\vec{d}}_{⊥}^{2} = {\vec{d}}_{1}^{2} - d_{▯} d_{⊥} = \sqrt{[{(4.0)}^{2} + {(5.0)}^{2} + {(- 6.0)}^{2} - {(- 3.2)}^{2}]} = 8.2 m$

Therefore, the component of ${\vec{d}}_{1}$ in the perpendicular direction of ${\vec{d}}_{2}$ is 8.2 m

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

Step 3: (a) Calculate the displacement r→

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

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

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

Step 3: (a) Calculate the displacement r→

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

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

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

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