Chapter 9.1, Problem 2E

1. (a) The length of a vector w = 〈a₁, a₂〉 is | w | = _____, so the length of the vector u in Figure II is

$\left|u\right|=\_\_\_\_\_.$

1. (b) If we know the length | w | and direction θ of a vector w, then we can express the vector in component form as

$w=\langle \boxed{},\boxed{}\rangle .$

To determine

To evaluate: The length of the vector $w$ and the length of the vector $u$.

Answer to Problem 2E

The length of the vector $w$ is $\sqrt{a^2+b^2}$ and the length of the vector $u$ is $2\sqrt{2}$.

Explanation of Solution

Formula used:

The formula to calculate the length of the vector $u=\langle a_1,a_2\rangle$ is,

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

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

$u=\langle x_2-x_1,y_2-y_1\rangle$ (2)

Calculation:

Substitute $a$ for $a_1$, $b$ for $a_2$ and $w$ for $u$ in equation (1) to calculate the length of the vector $w$.

$\left|w\right|=\sqrt{a^2+b^2}$

Hence, the length of the vector $w$ is $\sqrt{a^2+b^2}$.

The vectors $u$ and $v$ are shown in figure given below.

Figure (1)

Figure (1) shows the initial and terminal points of the vector $u$ are $\left(2,1\right)$ and $\left(4,3\right)$.

Substitute 2 for $x_1$, 4 for $x_2$, 1 for $y_1$ and 3 for $y_2$ in equation (2) to calculate $u$.

$\begin{array}{c}u=\langle 4-2,3-1\rangle \\ =\langle 2,2\rangle \end{array}$

In component form the value of $u$ is $\langle 2,2\rangle$.

Substitute 2 for $a_1$ and 2 for $a_2$ in equation (1) to calculate the length of the vector $u$.

$\begin{array}{c}\left|u\right|=\sqrt{2^2+2^2}\\ =\sqrt{4+4}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

Hence, the length of the vector $u$ is $2\sqrt{2}$.

To determine

To fill: The component form of the vector $w$.

Answer to Problem 2E

The component form of the vector $w$ is $\langle \left|w\right|\cos \theta \widehat{i},\left|w\right|\sin \theta \widehat{j}\rangle$.

Explanation of Solution

The formula to calculate the horizontal and vertical component of the vector $u$ with magnitude $\left|u\right|$ and direction $\theta$ is,

$\begin{array}{c}u=\langle a_1,a_2\rangle \\ =a_1\widehat{i}+a_2\widehat{j}\end{array}$

Where $a_1=\left|u\right|\cos \theta$ and $a_2=\left|u\right|\sin \theta$

Then, $u=\left|u\right|\cos \theta \widehat{i}+\left|u\right|\sin \theta \widehat{j}$ (1)

Substitute $w$ for $u$ in equation (1) to calculate the component form of the vector $w$.

$w=\left|w\right|\cos \theta \widehat{i}+\left|w\right|\sin \theta \widehat{j}$

Hence, the component form of the vector $w$ is $\langle \left|w\right|\cos \theta \widehat{i},\left|w\right|\sin \theta \widehat{j}\rangle$.