Q44E - University Physics with Modern Physics

Question

Find the vector product $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}\mathbf{\times }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$ (expressed in unit vectors) of the two vectors given in Exercise 1.38. What is the magnitude of the vector product?

Step-by-Step Solution

Answer

The magnitude of $\left|\stackrel{\rightharpoonup }{A}\right|=8.06,\left|\stackrel{\rightharpoonup }{B}\right|=5.38$.

Step 1: Given data

$\stackrel{\rightharpoonup }{A}= 4.00\hat{\mathit{\iota }}$ $+ 7.00\hat{J}$

$\stackrel{\rightharpoonup }{B }=5.00\widehat{\iota } - 2.00\hat{J}$

           

 

Step 2: The vector product of A → × B →

   $\stackrel{\rightharpoonup }{A}$$\times \stackrel{\rightharpoonup }{B}$ $\left[\begin{array}{c}i\\ 4\\ 5\end{array}\right.$$\begin{array}{l}j\\ 7\\ -2\end{array}$$\left.\begin{array}{c}\kappa \\ 0\\ 0\end{array}\right]$

$=\hat{\iota }\left(0-0\right)-\hat{J}\left(0-0\right)+\hat{\kappa }\left(-8-35\right)$

$=0\hat{I}+0\hat{J}-43\hat{\kappa }$

$= -43\hat{\kappa }$

                       

Step 3: The magnitude of the vector product

 

             The magnitude of vector is,

$\left|\stackrel{\rightharpoonup }{A}\right|$ $=\sqrt{4^{2 }+7^2}$$=\sqrt{65}$$=8.06$

 $\left|\stackrel{\rightharpoonup }{B}\right|$$=\sqrt{5^{2 }+2^2}$$=\sqrt{29}$$=5.3$