Question

The x and y components of four vectors, $\overline{a}$,$\overrightarrow{b}$,$\overrightarrow{c}$and $\overrightarrow{d}$are given below. For which vectors will your calculator give you the correct angle$\theta$when you use it tofind $\theta$with Eq. 3-6? Answer first by examining Fig. 3-12, and then check youranswers with your calculator.

$$\begin{gathered} a_x=3a_y=3c_x=-3c_y=-3 \\ {b}_x=-3b_y=3d_x=3d_y=-3 \end{gathered}$$

Short answer

for all vectors, the calculator gives you the correct angle $\theta$when you use it to find$\theta$

Step-by-step solution

Given information:

$$\begin{gathered} a_x=\text{3}{a}_y=\text{3}{c}_x=-\text{3}{c}_y=-\text{3} \\ {b}_x=-\text{3}{b}_y=\text{3}{d}_x=\text{3}{d}_y=-\text{3} \end{gathered}$$

Understanding the given information

The problem involves a simple calculation of the direction of the vector. A vector's orientation, or the angle it forms with the x-axis, determines its direction. A line with an arrow on top and a fixed point at the other end is used to represent a vector. The vector's direction can be determined by looking at the direction of its arrowhead.

Explanation for correct angle

The angle theta is defined as the tan inverse of the ratio of the y component to the x component of the vector.

For the vector $\overrightarrow{a}$,

$\theta ={\tan }^{-1}\frac{a_y}{a_x}$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{3}{3} \\ ={\tan }^{-1}\left(1\right) \\ ={45}^{\text{o}} \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of${\tan }^{-1}\left(1\right)$ which is angle 45^o fits the calculator range.

For the vector $\overrightarrow{b}$,

$\theta ={\tan }^{-1}\frac{b_y}{b_x}$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{3}{-3} \\ ={\tan }^{-1}\left(-1\right) \\ =-{45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of${\tan }^{-1}\left(1\right)$which is angle - 45^o fits the calculator range.

For the vector $\overrightarrow{c}$,

$\theta ={\tan }^{-1}\frac{c_y}{c_x}$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{-3}{-3} \\ ={\tan }^{-1}\left(1\right) \\ ={45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of${\tan }^{-1}\left(1\right)$which is angle 45^o fits the calculator range.

For vector 4,

$\theta ={\tan }^{-1}\frac{d_y}{d_x}$

$$\begin{gathered} \theta ={\tan }^{-1}\frac{-3}{3} \\ ={\tan }^{-1}\left(-1\right) \\ =-{45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of${\tan }^{-1}\left(1\right)$which is angle - 45^o fits the calculator range.

Thus, from the figure, it can be said that for all vectors the calculator give you the correct angle when you use it to find$\theta$