Chapter 10, Problem 1RE

(a)

To determine

To find: The component form of the vector $3u-4v$ .

Answer to Problem 1RE

The component form of the vector $3u-4v$ is $〈-17,32〉$ .

Explanation of Solution

Given information:

The given vectors are $u=〈-3,4〉$ and $v=〈2,-5〉$ .

Formula used:

Vector Addition:

Let $u=〈u_1,u_2〉$ and $v=〈v_1,v_2〉$ be vectors, then the sum/ resultant of the vectors $u$ and $v$ is given by:

  $\begin{array}{c}u+v=〈u_1,u_2〉+〈v_1,v_2〉\\ =〈u_1+v_1,u_2+v_2〉\end{array}$

Scalar multiplication:

The product of the scalar $k$ and the vector $u=〈u_1,u_2〉$ is given by:

  $\begin{array}{c}ku=k〈u_1,u_2〉\\ =〈k{u}_1,k{u}_2〉\end{array}$ .

Component form of a vector:

A two-dimensional vector $v$ is an ordered pair of real numbers, denoted in component form as

  $〈a,b〉$ .The numbers $a$ and $b$ are the components of the vector $v$ .

Calculation:

Substitute $u=〈-3,4〉$ and apply the formula for scalar multiplication of vectors to find $3u$ :

  $\begin{array}{c}3u=3〈-3,4〉\\ =〈3\left(-3\right),3\left(4\right)〉\\ =〈-9,12〉\end{array}$

Substitute $v=〈2,-5〉$ and apply the formula for scalar multiplication of vectors to find $4v$ :

  $\begin{array}{c}4v=4〈2,-5〉\\ =〈4\left(2\right),4\left(-5\right)〉\\ =〈8,-20〉\end{array}$

Substitute $3u=〈-9,12〉$ and $4v=〈8,-20〉$ , and apply the formula for vector addition to find $3u-4v$ :

  $\begin{array}{c}3u-4v=〈-9,12〉-〈8,-20〉\\ =〈-9-8,12-\left(-20\right)〉\\ =〈-17,32〉\end{array}$

According to the Component form of vectors,

The component form of the vector $3u-4v$ is $〈-17,32〉$ .

b)

To determine

To find: The magnitude of the vector $3u-4v$ .

Answer to Problem 1RE

The magnitude of the vector $3u-4v$ is $36.23$ .

Explanation of Solution

Given information:The given vectors are $u=〈-3,4〉$ and $v=〈2,-5〉$ .

Formula used:

Magnitude of a vector:

The magnitude of the vector $〈a,b〉$ is the non-negative real number:

  $\left|〈a,b〉\right|=\sqrt{a^2+b^2}$ .

Vector Addition:

Let $u=〈u_1,u_2〉$ and $v=〈v_1,v_2〉$ be vectors, then the sum/ resultant of the vectors $u$ and $v$ is given by:

  $\begin{array}{c}u+v=〈u_1,u_2〉+〈v_1,v_2〉\\ =〈u_1+v_1,u_2+v_2〉\end{array}$

Scalar multiplication:

The product of the scalar $k$ and the vector $u=〈u_1,u_2〉$ is given by:

  $\begin{array}{c}ku=k〈u_1,u_2〉\\ =〈k{u}_1,k{u}_2〉\end{array}$ .

Calculation:

Substitute $u=〈-3,4〉$ and apply the formula for scalar multiplication of vectors to find $3u$ :

  $\begin{array}{c}3u=3〈-3,4〉\\ =〈3\left(-3\right),3\left(4\right)〉\\ =〈-9,12〉\end{array}$

Substitute $v=〈2,-5〉$ and apply the formula for scalar multiplication of vectors to find $4v$ :

  $\begin{array}{c}4v=4〈2,-5〉\\ =〈4\left(2\right),4\left(-5\right)〉\\ =〈8,-20〉\end{array}$

Substitute $3u=〈-9,12〉$ and $4v=〈8,-20〉$ , and apply the formula for vector addition to find $3u-4v$ :

  $\begin{array}{c}3u-4v=〈-9,12〉-〈8,-20〉\\ =〈-9-8,12-\left(-20\right)〉\\ =〈-17,32〉\end{array}$

Therefore, the vector $3u-4v$ is $〈-17,32〉$ .

Substitute $3u-4v=〈-17,32〉$ and apply the formula for magnitude of vectors to find the value of $\left|3u-4v\right|$ :

  $\begin{array}{c}\left|3u-4v\right|=\left|〈-17,32〉\right|\\ =\sqrt{{\left(-17\right)}^2+{32}^2}\\ =\sqrt{289+1024}\\ =\sqrt{1313}\\ \approx 36.23\end{array}$

Thus, the magnitude of the vector $3u-4v$ is $36.23$ .