Q45E - University Physics with Modern Physics

Question

Find the angle between each of these pairs of vectors:

(a) $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{ }\mathbf{ }\mathbf{and}\mathbf{ }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}$ 

(b)$\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\widehat{\mathbf{j}\mathbf{ }}\mathbf{ }\mathbf{and}\mathbf{ }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}$

(c) $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\widehat{\mathbf{j}\mathbf{ }}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{ }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{14}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}$ 

Step-by-Step Solution

Answer

The angles are ${\cos }^{-1}\left(\frac{-22}{\sqrt{40}\sqrt{13}}\right),{\cos }^{-1}\left(\frac{60}{\sqrt{34}\sqrt{136}}\right)$ and $0$ .

Given data

    $\overrightarrow{A}=-2.00\hat{i}+6.00\hat{j}$ and $\overrightarrow{B}=2.00\hat{i}-3.00\hat{j}$

   $\overrightarrow{A}=3.00\hat{i}+5.00\hat{j}$ and $\overrightarrow{B}=10.00\hat{i}+6.00\hat{j}$

   $\overrightarrow{A}=-4.00\hat{i}+2.00\hat{j}$ and $\overrightarrow{B}=7.00\hat{i}+14.00\hat{j}$

( a ) The   angle   between A   →   and   B →

   $$\begin{gathered} \overrightarrow{A}\overrightarrow{B}=(-2\times 2)+(6\times (-3\left)\right)=-22 \\ A=\sqrt{2^2+6^2}=\sqrt{40}=6.32 \\ B=\sqrt{2^2+3^2}=\sqrt{13}=3.6 \end{gathered}$$

   $$\begin{gathered} \cos \theta =\frac{\overrightarrow{A}·\overrightarrow{B}}{AB}=\frac{60}{\sqrt{34}\sqrt{136}} \\ \theta ={\cos }^{-1}\left(\frac{60}{\sqrt{34}\sqrt{136}}\right) \end{gathered}$$

    ( b )   The   angle   between   A →   and   B →

    

      $$\begin{gathered} \overrightarrow{A}\overrightarrow{B}=(-4\times 7)+(2\times 14)=0 \\ A=\sqrt{4^2+2^2}=\sqrt{20} \\ B=\sqrt{7^2+{14}^2}=\sqrt{245} \\ \cos \theta =\frac{\overrightarrow{A}·\overrightarrow{B}}{AB}=\frac{-22}{\sqrt{40}\sqrt{13}} \\ \theta ={\cos }^{-1}\left(\frac{-22}{\sqrt{40}\sqrt{13}}\right) \end{gathered}$$

( c )   The   angle   between   A →   and   B →

    

       $$\begin{gathered} \overrightarrow{A}\overrightarrow{B}=(-4\times 7)+(2\times 14)=0 \\ A=\sqrt{4^2+2^2}=\sqrt{20} \\ B=\sqrt{7^2+{14}^2}=\sqrt{245} \\ \cos \theta =\frac{\overrightarrow{A}·\overrightarrow{B}}{AB}=\frac{0}{\sqrt{20}\sqrt{245}}=0 \end{gathered}$$