Chapter 3, Problem 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.

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.

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 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 $5.00\widehat{i}+4.00\widehat{j}$

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 $-1.00\widehat{i}+8.00\widehat{j}$.

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 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 $\left(6.4,38.65°\right)$.

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 $\left(8.06,97.13°\right)$.

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.