Chapter 6, Problem 12RE

Solution

a)

To find: A unit vector in the direction of $\overrightarrow{AB}$ .

The unit vector in the direction of $\overrightarrow{AB}$ is $〈1,0〉.$

Given information:

Given two points $A=\left(3,1\right),B=\left(5,1\right)$ .

Formula used:

• The length of the vector $v=〈x,y〉$ is $\sqrt{x^2+y^2}$ , and denoted as $\left|v\right|.$
• The unit vector in the direction of $v$ is $\frac{1}{\left|v\right|}v.$
• If $P=\left(x,y\right)$ is a point, then the vector $\overrightarrow{OP}$ is starting at the origin $O$ and ends with the point $P$ . The vector $\overrightarrow{OP}$ is written as $〈x,y〉.$
• Head Minus Tail (HMT) Rule: Let $P=\left(x_1,y_1\right),Q=\left(x_2,y_2\right)$ be two points. The vector $\overrightarrow{PQ}$ is represented as $〈x_2-x_1,y_2-y_1〉.$

Calculation:

The vector $\overrightarrow{AB}$ is computed as follows:

  $\begin{array}{c}\overrightarrow{AB}=〈5-3,1-1〉\\ =〈2,0〉.\end{array}$

The vector $\overrightarrow{AB}$ is $〈2,0〉.$

Substitute 2 for $x$ and $0$ for $y$ in $\sqrt{x^2+y^2}$ , the length of the vector $〈2,0〉$ is $\begin{array}{c}\sqrt{{\left(2\right)}^2+{\left(0\right)}^2}=\sqrt{4}\\ =2.\end{array}$

Thus, the length of the vector $\overrightarrow{AB}$ is $\left|\overrightarrow{AB}\right|=2.$

The unit vector in the direction of $\overrightarrow{AB}$ is computed as follows:

  $\begin{array}{c}\frac{1}{\left|\overrightarrow{AB}\right|}\overrightarrow{AB}=\frac{1}{2}〈2,0〉\\ =〈1,0〉.\end{array}$

Thus, the unit vector in the direction of $\overrightarrow{AB}$ is $〈1,0〉.$

b)

To find: A vector of magnitude 3 in the opposite direction of $\overrightarrow{AB}$ .

Vector of magnitude 3 in the opposite direction of $\overrightarrow{AB}$ is $〈-3,0〉.$

Given information:

Given two points $A=\left(3,1\right),B=\left(5,1\right)$ .

From part (a), the unit vector in the direction of $\overrightarrow{AB}$ is $〈1,0〉.$

Formula used:

• If $u$ is a unit vector, then the vector of length $\lambda$ in the opposite direction of $u$ is $-\lambda u.$

Calculation:

Write $u=\frac{\overrightarrow{AB}}{\left|\overrightarrow{AB}\right|}.$

A vector of magnitude 3 in the opposite direction of $\overrightarrow{AB}$ is $-3u.$

Substitute $〈1,0〉$ for $u,$ then

  $\begin{array}{c}-3u=-3〈1,0〉\\ =〈-3,0〉.\end{array}$

Therefore, vector of magnitude 3 in the opposite direction of $\overrightarrow{AB}$ is $〈-3,0〉.$