Q47E - University Physics with Modern Physics

Question

For the two vectors $\overrightarrow{\mathbf{A}}$ an $\overrightarrow{\mathbf{D}}$ in Fig. E1.24, find the magnitude and direction of (a) the vector product $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{D}}$ ; (b) the vector product $\overrightarrow{\mathbf{D}}\mathbf{\times }\overrightarrow{\mathbf{A}}$

Step-by-Step Solution

Answer

The magnitude and direction are $\overrightarrow{A}\times \overrightarrow{D}=63.8908408, -\hat{\kappa }$  and  $\overrightarrow{D}\times \overrightarrow{A}$ = $63.8908408$ ,$\stackrel{⏜}{\mathrm{K}}$ .

Step 1: Given data

$$\begin{gathered} \overrightarrow{A}=8.00m \\ \overrightarrow{D}= 10.0m \end{gathered}$$

Step 2: Solution

        $\overrightarrow{A}=8\times \cos 270° \widehat{\iota }$$+8\times \sin 270° \hat{j}$$=-8\hat{j}$

      $\overrightarrow{D }=10\times \cos 143° \hat{\iota }$$+10\times \sin 143°\hat{j}$

          $= -7.9863551\hat{i}$$+6.08150232\hat{j}$

Step 3: The vector product

$\overrightarrow{A}\times \overrightarrow{D}$

$$\begin{gathered} =\left(-8\hat{j}\right)\times \left(-7.9863551\hat{i}+6.08150232\hat{j}\right) \\ =+\left(8\times 7.9863551\left(\hat{j}\times \hat{i}\right)\right) \\ =-63.8908408\hat{\kappa } \end{gathered}$$

$\mathrm{Magnitude} = 63.8908408$

$\mathrm{Direction} = -\stackrel{⏜}{\mathrm{k}}$     

Step 4: The vector product D → × A   →                 D → × A   →                 D → × A   →        

$$\begin{gathered} \overrightarrow{D}\times \overrightarrow{A} \\ =\left(-7.9863551\hat{i}+6.08150232\hat{j}\right)\times \left(-8\hat{j}\right) \\ =+63.8908408\hat{\kappa } \end{gathered}$$

       $\mathrm{Magnitude} = 63.8908408$ 

      $\mathrm{Direction} = \stackrel{⏜}{\mathrm{k}}$