Chapter 3, Problem 3.26P

Given the vectors $\stackrel{\to }{A}=2.00\hat{i}+6.00\hat{j}$ and $\stackrel{\to }{B}=3.00\hat{i}-2.00\hat{j}$, (a) draw the vector sum $\stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}$ and the vector difference $\stackrel{\to }{D}=\stackrel{\to }{A}-\stackrel{\to }{B}$. (b) Calculate $\stackrel{\to }{C}$ and $\stackrel{\to }{D}$, in terms of unit vectors, (c) Calculate $\stackrel{\to }{C}$ and $\stackrel{\to }{D}$ in terms of polar coordinates, with angles measured with respect to the positive x axis.

To determine

To sketch: The diagram for the vector sum $\overrightarrow{A}+\overrightarrow{B}$ and the vector difference $\overrightarrow{A}-\overrightarrow{B}$.

Explanation of Solution

Introduction: Graphical method or geographical construction of vectors is defined as a method that is used to add more than two vectors. The resultant vector is drawn from the tail of the first vector to the head of the last vector.

Section 1:

To sketch: The diagram for the vector sum, $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$

Introduction: Graphical method or geographical construction of vectors is defined as a method that is used to add more than two vectors. The resultant vector is drawn from the tail of the first vector to the head of the last vector.

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

Draw the vector diagram for vector sum, $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$.

Figure I

Conclusion:

Therefore, the vector diagram for vector sum, $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ is drawn.

Section 2:

To sketch: The diagram for the vector difference, $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$.

Introduction: Graphical method or geographical construction of vectors is defined as a method that is used to add more than two vectors. The resultant vector is drawn from the tail of the first vector to the head of the last vector.

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

Draw the vector diagram for vector difference, $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$.

Figure II

Conclusion:

Therefore, the vector diagram for vector difference, $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$ is drawn.

To determine

The vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of unit vectors.

Answer to Problem 3.26P

The vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of unit vectors are equal to $5.00\widehat{i}+4.00\widehat{j}$ and $-1.00\widehat{i}+8.00\widehat{j}$ respectively.

Explanation of Solution

Section 1:

To determine: The vector $\overrightarrow{C}$ in terms of unit vectors.

Answer: The vector $\overrightarrow{C}$ in terms of unit vectors is equals to $5.00\widehat{i}+4.00\widehat{j}$

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

Formula to calculate the $\overrightarrow{C}$ in terms of unit vectors is,

$\overrightarrow{C}=\left(A_{\text{x}}+B_{\text{x}}\right)\widehat{i}+\left(A_{\text{y}}+B_{\text{y}}\right)\widehat{j}$

Substitute $2.00$ for $A_{\text{x}}$, $6.00$ for $A_{\text{y}}$, $3.00$ for $B_{\text{x}}$ and $-2.00$ for $B_{\text{y}}$ in above expression.

$\begin{array}{c}\overrightarrow{C}=\left(2.00+3.00\right)\widehat{i}+\left(6.00+\left(-2.00\right)\right)\widehat{j}\\ =5.00\widehat{i}+4.00\widehat{j}\end{array}$

Conclusion:

Therefore, the vector $\overrightarrow{C}$ in terms of unit vectors is equals to $5.00\widehat{i}+4.00\widehat{j}$.

Section 2:

To determine: The vector $\overrightarrow{D}$ in terms of unit vectors.

Answer: The vector $\overrightarrow{D}$ in terms of unit vectors is equals to $-1.00\widehat{i}+8.00\widehat{j}$.

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

Formula to calculate the $\overrightarrow{D}$ in terms of unit vectors is,

$\overrightarrow{D}=\left(A_{\text{x}}-B_{\text{x}}\right)\widehat{i}+\left(A_{\text{y}}-B_{\text{y}}\right)\widehat{j}$

Substitute $2.00$ for $A_{\text{x}}$, $6.00$ for $A_{\text{y}}$, $3.00$ for $B_{\text{x}}$ and $-2.00$ for $B_{\text{y}}$ in above expression.

$\begin{array}{c}\overrightarrow{D}=\left(2.00-3.00\right)\widehat{i}+\left(6.00-\left(-2.00\right)\right)\widehat{j}\\ =-1.00\widehat{i}+8.00\widehat{j}\end{array}$

Conclusion:

Therefore, the vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of unit vectors are equal to $5.00\widehat{i}+4.00\widehat{j}$ and $-1.00\widehat{i}+8.00\widehat{j}$ respectively.

To determine

The vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of polar coordinates.

Answer to Problem 3.26P

The vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of polar coordinates are equal to $\left(6.4,38.65°\right)$ and $\left(8.06,97.13°\right)$ respectively.

Explanation of Solution

Section 1:

To determine: The vector $\overrightarrow{C}$ in terms of polar coordinates.

Answer: The vector $\overrightarrow{C}$ in terms of polar coordinates is equals to $\left(6.4,38.65°\right)$.

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

The vector $\overrightarrow{C}$ in terms of unit vector is given as,

$\overrightarrow{C}=5.00\widehat{i}+4.00\widehat{j}$

Formula to calculate the first polar coordinate is,

$r=\sqrt{C_{\text{x}}^2+C_{\text{y}}^2}$

Substitute $5.00$ for $C_{\text{x}}$ and $4.00$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}r=\sqrt{{\left(5.00\right)}^2+{\left(4.00\right)}^2}\\ =6.4\end{array}$

Formula to calculate the second polar coordinate is,

$\theta ={\tan }^{-1}\left(\frac{C_{\text{y}}}{C_{\text{x}}}\right)$

Substitute $5.00$ for $C_{\text{x}}$ and $4.00$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{4.00}{5.00}\right)\\ =38.65°\end{array}$

Conclusion:

Therefore, the vector $\overrightarrow{C}$ in terms of polar coordinates is equals to $\left(6.4,38.65°\right)$.

Section 2:

To determine: The vector $\overrightarrow{D}$ in terms of polar coordinates.

Answer: The vector $\overrightarrow{D}$ in terms of polar coordinates is equals to $\left(8.06,97.13°\right)$.

Explanation:

Given information:

Vector $\overrightarrow{A}$ in terms of unit vectors is equal to $2.00\widehat{i}+6.00\widehat{j}$ and vector $\overrightarrow{B}$ in terms of unit vectors is equal to $3.00\widehat{i}-2.00\widehat{j}$.

The vector $\overrightarrow{D}$ in terms of unit vector is given as,

$\overrightarrow{D}=-1.00\widehat{i}+8.00\widehat{j}$

Formula to calculate the first polar coordinate is,

$r=\sqrt{D_{\text{x}}^2+D_{\text{y}}^2}$

Substitute $-1.00$ for $D_{\text{x}}$ and $8.00$ for $D_{\text{y}}$ in above expression.

$\begin{array}{c}r=\sqrt{{\left(-1.00\right)}^2+{\left(8.00\right)}^2}\\ =8.06\end{array}$

Formula to calculate the second polar coordinate is,

$\theta ={\tan }^{-1}\left(\frac{C_{\text{y}}}{C_{\text{x}}}\right)$

Substitute $-1.00$ for $D_{\text{x}}$ and $8.00$ for $D_{\text{y}}$ in above expression.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{8.00}{-1.00}\right)\\ =97.13°\end{array}$

Conclusion:

Therefore, the vectors $\overrightarrow{C}$ and $\overrightarrow{D}$ in terms of polar coordinates are equal to $\left(6.4,38.65°\right)$ and $\left(8.06,97.13°\right)$ respectively.