Chapter 10.2, Problem 5E

To determine

To find: The magnitude of the vector and the direction of angle $\theta$ .

Answer to Problem 5E

The answer:

The magnitude of the vector $|(2,2)|=2\sqrt{2}$

And

The angle $\theta ={45}^{\circ }$

Explanation of Solution

Given information:

The positive $x$ -axis $\left(0\le \theta <{360}^{\circ }\right)$ .

The vector $(2,2)$ .

Calculation:

Know that the magnitude of a vector $(a,b)$ is equal to:

  $|(a,b)|=\sqrt{a^2+b^2}$

Know that for the direction angle $\theta$ of a vector $(a,b)$ with the positive $x$ -axis,

  $\begin{array}{c}\sin \theta =\frac{b}{|(a,b)|}\\ \cos \theta =\frac{a}{|(a,b)|}\\ \Rightarrow \tan \theta =\frac{b}{a}\end{array}$

To use the previous results, the magnitude of the vector $(2,2)$ is equal to:

  $\begin{array}{c}|(2,2)|=\sqrt{2^2+2^2}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

While for the direction angle $\theta$ of a vector $(2,2)$ with the positive $x$ -axis,

  $\begin{array}{c}\tan \theta =\frac{2}{2}\\ =1\\ \Rightarrow \theta ={45}^{\circ }\end{array}$

Because,

  $\begin{array}{c}\sin \theta >0\\ \cos \theta >0\end{array}$

Therefore, the required magnitude of the vector is $|(2,2)|=2\sqrt{2}$ and the direction of angle $\theta$ is

  $\theta ={45}^{\circ }$ .