Q 28E - University Physics with Modern Physics

Question

Let  $\mathbf{\theta }$ be the angle that the vector  $\overrightarrow{\mathbf{A}}$ makes with the +x-axis, measured counterclockwise from that axis. Find angle $\mathbf{\theta }$  for a vector that has these components: (a) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}\mathbf{ }{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$ ; (b) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$ ; (c) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$ ; (d)  ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{m}$. 

Step-by-Step Solution

Answer

The angle made by the vectors can be summarised as, 

(a)  333.$4°$,

(b)  $26.6°$

(c)  $333.4°$ and

(d)  $26.6°$

Step 1: Angle between two vectors

Here different coordinates of the vector are given. We are knowing that in coordinates x component of the vector will be first and then the y component. Now to find the angle of the vector with respect to the x-axis we just need to take the ratio of the y component to that of the x component and then take the tan inverse of that value.

 

This will give you the angle of the vector with the x-axis. 

 

It can be represented as 

 

$\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right)$ …………………(1)

Step 2: (a) Calculation of the θ for A x = 2 . 00   m , A y = - 1 . 00   m

Substitute the given data in equation (1), and we get,

 $$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00m}{2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$

Step 3: (b) Calculation of the θ for A x = 2 . 00   m , A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

 

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{1.00m}{2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.6° \end{gathered}$$ 

Step 4: (c) Calculation of the θ for A x = - 2 . 00   m , A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

 

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00m}{2.00m}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$ 

Step 5: (d) Calculation of the θ for A x = - 2 . 00   m , A y = - 1 . 00   m

Substitute the given data in equation (1), and we get,

 

 $$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 m}{-2.00 m}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.6° \end{gathered}$$

 

Therefore the angle made by the vectors is (a) $333.4°$ , (b) $26.6°$ , (c) $333.4°$ , and

(d) $26.6°$ .