Q45 E - University Physics with Modern Physics

Question

 Find the angle between each of these pairs of vectors:

(a) $\overrightarrow{A }=-2.00i \mathit{+}\mathit{ }\mathit{6}\mathit{.}\mathit{00}j$ and $\overrightarrow{B}=2.00\hat{i}$$-3.00\hat{j}$

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

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

Step 1: Given data

 $\stackrel{\rightharpoonup }{A}$$=-2.00\hat{\iota }$$+6.00\hat{J}$ and  $\stackrel{\rightharpoonup }{B}$$=2.00\hat{\iota }$$-3.00\hat{J}$

                         

  $\stackrel{\rightharpoonup }{A}$$=3.00\hat{\iota }$$+5.00\hat{J}$ and $\overrightarrow{B}$$=10.00\hat{\iota }$ + data-custom-editor="chemistry" $6.00\hat{\mathrm{J}}$

 $\overrightarrow{A}$$=-4.00\hat{\iota }$$+2.00\hat{J}$ and  $\stackrel{\rightharpoonup }{B}$$= 7.00\hat{\iota }$$+14.00\hat{J}$

Step 2: (a) The angle between A → and B →

        $\overrightarrow{A}·\overrightarrow{B}$$=\left(-2\times 2\right)+\left(6\times \left(-3\right)\right)=-22$

         $A=\sqrt{2^{2 }+6^2}$$=\sqrt{40}$$=6.32$

          $B=\sqrt{2^2+3^2}$$=\sqrt{13}$$=3.6$  

$\cos \theta =\frac{\overrightarrow{A }.\overrightarrow{B}}{AB}=\frac{60}{\sqrt{34}\sqrt{136}}$

           $\theta ={\cos }^{-1}\left(\frac{60}{\sqrt{34}\sqrt{136}}\right)$

     

 

               

 

                 

Step 2: (b) The angle between A → and B →

$\overrightarrow{A}.\overrightarrow{B}=\left(-4\times 7\right)+\left(2\times 14\right)=0$

$A=\sqrt{4^2+2^2}=\sqrt{20}$

$B=\sqrt{7^2+{14}^2}=\sqrt{245}$

$\cos \theta = \frac{\overrightarrow{A} . \overrightarrow{B}}{AB}=\frac{-22}{\sqrt{40}\sqrt{13}}$

$\theta ={\cos }^{-1}\left(\frac{-22}{\sqrt{40}\sqrt{13}}\right)$

Step 3: ( a) The angle between A → and B →

$\overrightarrow{A}.\overrightarrow{B}$$=\left(-4\times 7\right)+\left(2\times 14\right)=0$

 $$\begin{gathered} A={\sqrt{4^2+2^2}}^{}=\sqrt{20} \\ B= \sqrt{7^2+{14}^2}=\sqrt{245} \end{gathered}$$

$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{AB}=\frac{0}{\sqrt{20}\sqrt{245}}=0$