Chapter 9, Problem 1RCC

(a)

To determine

To explain: The vector in a plane and representation of vector in the coordinate plane.

Explanation of Solution

A line segment with a particular direction running from initial point to terminal point is called a vector in the plane. A vector in the plane is denoted by $\overrightarrow{AB}$, where arrow specify the direction, A is the initial point and B is the terminal point of the vector $\overrightarrow{AB}$.

The vector $v$ as an ordered pair of real numbers in the coordinate plane is represented by,

$v=\langle a_1,a_2\rangle$

Where,

$a_1$ is the horizontal component of $v$

$a_2$ is the vertical component of $v$

A vector represents a magnitude and a direction.

(b)

To determine

To find: The vector with initial point $\left(2,3\right)$ and terminal point $\left(4,10\right)$.

Answer to Problem 1RCC

The vector with initial point $\left(2,3\right)$ and terminal point $\left(4,10\right)$ is $\langle 2,7\rangle$.

Explanation of Solution

Given:

The initial point is $\left(2,3\right)$ and terminal point is $\left(4,10\right)$.

Formula used:

The formula to calculate the vector $v$ in the plane with initial point $P\left(x_1,y_1\right)$ and terminal point $Q\left(x_2,y_2\right)$ is,

$v=\langle x_2-x_1,y_2-y_1\rangle$ (1)

Calculation:

Substitute 2 for $x_1$, 4 for $x_2$,3 for $y_1$,10 for $y_2$ in equation (1).

$\begin{array}{c}v=\langle 4-2,10-3\rangle \\ v=\langle 2,7\rangle \end{array}$

Thus, the vector with initial point $\left(2,3\right)$ and terminal point $\left(4,10\right)$ is $\langle 2,7\rangle$.

(c)

To determine

The terminal point of the vector $v$ and sketch its several representations.

Answer to Problem 1RCC

The terminal point of the vector $v$ is $\langle 3,2\rangle$ as shown in the Figure (1).

Explanation of Solution

Given:

The vector $v=\langle 2,1\rangle$ and the initial point is $P\left(1,1\right)$.

Calculation:

Section (a):

The terminal point of $v$ be $\left(x,y\right)$.

Substitute x for $x_2$, 1 for $x_1$, y for $y_2$,1 for $y_1$, $\langle 2,1\rangle$ for $v$ in equation (1).

$\langle 2,1\rangle =\langle x-1,y-1\rangle$

Compare both sides,

$\begin{array}{c}x-1=2\text{and}y-1=1\\ x=2+1\text{and}y=1+1\\ x=3\text{and}y=2\end{array}$

Thus, the terminal point of the vector $v$ is $\langle 3,2\rangle$.

Section (b):

Draw the graph of vector as shown below,

Figure (1)

Thus, Figure (1) shows various representations of the vector $v$ with several initial points.

(d)

To determine

The definition of magnitude of vector and the value of the vector $w=\langle 3,4\rangle$.

Explanation of Solution

Calculation:

The length of the line segment is called the magnitude of the vector and it is denoted by $\left|\overrightarrow{AB}\right|$. The magnitude of a vector $w=\langle a_1,a_2\rangle$ is,

$\left|w\right|=\sqrt{a_1^2+a_2^2}$ (1)

The magnitude of the vector $w=\langle 3,4\rangle$ is calculated as,

$\begin{array}{c}\left|w\right|=\sqrt{3^2+4^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Thus, magnitude of vector $w=\langle 3,4\rangle$ is 5.

(e)

To determine

The vectors $i$ and $j$ and express the vector in terms of $i$ and $j$.

Explanation of Solution

Calculation:

A vector of length 1 is called a unit vector. The vectors $i$ and $j$ are two useful unit vectors defined by,

$v=\langle a_1,a_2\rangle =a_{1}i+a_{2}j$

The vector $v=\langle 5,9\rangle$ is expressed in terms of vectors $i$ and $j$ as,

$v=5i+9j$

Thus, the vector $v=\langle 5,9\rangle$ in terms of $i$ and $j$ is $v=5i+9j$.

(f)

To determine

The direction $\theta$ of vector $v$ and the coordinates of $v$ in terms of its length and direction and the figure to illustrate the answer.

Explanation of Solution

Calculation:

The smallest positive angle in standard position formed by the positive x-axis and the vector $v$ is known as the direction of $v$ and it is denoted by $\theta$.

The vector $v=\langle a_1,a_2\rangle$ has the coordinates in terms of length and direction as,

$a_1=\left|v\right|\cos \theta$ (1)

And

$a_2=\left|v\right|\sin \theta$ (2)

Where,

$a_1$ is the horizontal component

$a_2$ is the vertical component.

The vector $v$ is expressed as,

$v=\left|v\right|\cos \theta i+\left|v\right|\sin \theta j$ (5)

The graph for the above equation is,

Figure (2)

Thus, Figure (2) shows the graph of the coordinates of a vector in terms of length and direction.

(g)

To determine

To find: The vector $v$ in terms of its coordinates.

Answer to Problem 1RCC

The vector $v$ in terms of its coordinates is $v=\langle \frac{5\sqrt{3}}{2},\frac{5}{2}\rangle$.

Explanation of Solution

Given:

The length $\left|v\right|=5$ and direction $\theta =\pi /6$.

Calculation:

The vector $v=\langle a_1,a_2\rangle$ has the coordinates in terms of length and direction as,

$a_1=\left|v\right|\cos \theta$ (1)

And

$a_2=\left|v\right|\sin \theta$ (2)

Substitute 5 for $\left|v\right|$, $\pi /6$ for $\theta$ in equation (1) and equation (2).

$\begin{array}{c}{a}_1=5\times \cos \left(\pi /6\right)\\ =5\times \frac{\sqrt{3}}{2}\\ a_1=\frac{5\sqrt{3}}{2}\end{array}$

And

$\begin{array}{c}{a}_2=5\times \sin \left(\pi /6\right)\\ =5\times \frac{1}{2}\\ a_2=\frac{5}{2}\end{array}$

Substitute $\frac{5\sqrt{3}}{2}$ for $a_1$, $\frac{5}{2}$ for $a_2$ in $v=\langle a_1,a_2\rangle$.

$v=\langle \frac{5\sqrt{3}}{2},\frac{5}{2}\rangle$

Thus, the coordinates of the vector $v$ is $\langle \frac{5\sqrt{3}}{2},\frac{5}{2}\rangle$.